The Multi-Dimensional Frobenius Problem 2 Jeffrey Amos, Charles Chen, Olga Lepigina, Timur Nezhmetdinov, Darren Ong, Matthew Richer, Laura Zirbel Under the Guidance of Professor Vadim Ponomarenko Research Experiences for Undergraduates, Trinity University July 27, 2006 Abstract Consider the problem of determining maximal vectors g such that the Diophantine system Mx = g has no solution. We provide a variety of results to this end: conditions for the existence of g, conditions for the uniqueness of g, bounds on g, determining g explicitly in several important special cases, constructions for g, and a reduction for M . 1
Transcript
The Multi-Dimensional Frobenius Problem 2
Jeffrey Amos, Charles Chen, Olga Lepigina,Timur Nezhmetdinov, Darren Ong, Matthew Richer, Laura Zirbel
Under the Guidance of Professor Vadim PonomarenkoResearch Experiences for Undergraduates, Trinity University
July 27, 2006
Abstract
Consider the problem of determining maximal vectors g such thatthe Diophantine system Mx = g has no solution. We provide a varietyof results to this end: conditions for the existence of g, conditions forthe uniqueness of g, bounds on g, determining g explicitly in severalimportant special cases, constructions for g, and a reduction for M .
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1 Introduction
We will begin where last year’s combinatorics group stopped. We will beusing the notation used in the paper that was recently submitted for pub-lication. In fact, almost all of this introduction was copied straight from itwith the proofs removed.
Let m, x be column vectors from N0. Georg Frobenius focused atten-tion on determining maximal g such that the linear Diophantine equationmT x = g has no solutions. This problem has attracted substantial atten-tion in the last 100+ years; for a survey, see a really cool book that Vadimchecked out from the library, which contains almost 500 references as well asapplications to algebraic geometry, coding theory, linear algebra, algorithmanalysis, discrete distributed systems, and random vector generation. A nat-ural generalization of this problem (and essential to some applications) isto determine maximal vector(s) g such that the system of linear Diophan-tine equations Mx = g has no solutions. This has attracted relatively littleattention, perhaps because maximality must be subject to a partial vector or-dering. We attempt to redress this injustice by providing a variety of resultsin this multi-dimensional context.
We fix Rn. For any real matrix X and any S ⊆ R, we write XS forXs : s ∈ Sk, where k denotes the number of columns of X. Abusing thisnotation slightly, we write X1 for the vector X1k. We fix M ⊆ Zn×(n+m),and write M = [A|B], where A is n × n. We call AR≥0 the cone, and MN0
the monoid. |A| denotes henceforth the absolute value of det A. If |A| 6= 0,then we follow others and call the cone volume. If, in addition, each columnof B lies in the cone, then we call M simplicial. Unless otherwise noted, weassume henceforth that M is simplicial. Note that if n ≤ 2, then we mayalways rearrange columns to make M simplicial.
Let u, v ∈ Rn. If u − v ∈ AZ, then we write u ≡ v and say that u, v areequivalent mod A. If u − v ∈ AR≥0 , then we write u ≥ v. If u − v ∈ AR>0 ,then we write u v. Note that u v implies u ≥ v, and u v ≥ w impliesu w; however, u v does not necessarily imply that u v. For v ∈ Rn, wewrite [v] = u ∈ Zn : u v. We say that v is complete if [v] ⊆MN0 . Weset G, more precisely G(M), to be the set of all ≥-minimal complete vectors.We call elements of G Frobenius vectors ; they are the vector analogue of gthat we will investigate.
Set Q = (1/|A|)Zn ⊆ Q. Although G is defined in Rn, in fact it is asubset of Qn, by the following result. Furthermore, the columns of B are
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in AQ≥0 ; hence MQ≥0 = AQ≥0 and without loss we henceforth work over Qrather than over R.
Proposition 1.1. Let v ∈ Rn. There exists v? ∈ Qn with [ v] = [Av?]and v ≥ Av?.
In general, MN0 does not form an ≤-lattice, because A−1B does not haveinteger entries and thus lub is not well-defined. However, because
(Q≥0
)nis a
chain product, our partial order ≤ is a lattice over Q. For x = Ax′, y = Ay′,we see that lub(x, y) = Az′, where z′ is defined via (z′)i = max((x′)i, (y
′)i).For u ∈ Qn, we set V (u) =
(u + AQ∩(0,1]
)∩Zn. It was known to Dedekind
that |V (u)| = |A|, and that V (u) is a complete set of coset representativesmod A (as restricted to Zn).
The following equivalent conditions on M generalize the one-dimensionalnotion of relatively prime generators. We assume henceforth, unless other-wise noted, that M possesses these properties. We call such M dense.
Theorem 1.2. The following are equivalent:
1. G is nonempty.
2. MZ = Zn.
3. For all unit vectors ei (1 ≤ i ≤ n), ei ∈MZ.
4. There is some v ∈MN0 with v + ei ∈MN0 for all unit vectors ei.
5. The GCD of all the n× n minors of M has absolute value 1.
6. The elementary divisors of M are all 1.
Classically, there is a second type of Frobenius number f , maximal sothat mT x = f has no solutions with x from N (rather than N0). This doesnot add much; it was shown that f = g + mT 1. A similar situation holds inthe vector context.
Proposition 1.3. Call v f-complete if [v] ⊆MN. Set F to be all ≥-minimalf-complete vectors. Then F = G + M1.
For vector u and i ∈ [1, n], let Ci(u) = v : v ∈ Zn \ MN0 , v = u +Aw, (w)i = 0, (w)j ∈ (0, 1] for j 6= i and let C(u) =
⋃i∈[1,n] C
i(u), a disjoint
union. Call elements of C(u) critical. Note that if v ∈ Ci(u), then v + Aei ∈V (u). Critical elements characterize G, as shown by the following.
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Theorem 1.4. Let x be complete. x ∈ G if and only if Ci(x) 6= ∅, ∀i ∈ [1, n].
Theorem 1.5. Fix A and vector c ≥ 0. Set C = c(1n)T , a square matrix,and fix k ∈ N. Set M = [A|A + C|A + 2C| · · · |A + kC]. Suppose that M isdense. Then G(M) = Ax + |A|c− A1 − c : x ∈ Nn
0 , ‖x‖1 = d(|A| − 1)/ke.
Let MIN = x : x ∈ MN0 ; for all y ∈ MN0 , if y ≡ x then y ≥ x. Pro-vided M is dense, MIN will have at least one representative of each of the|A| equivalence classes mod A. MIN is a generalization of a one-dimensionalmethod; the following result shows that it characterizes G.
Theorem 1.6. Let g ∈ G. Then g = lub(N) − A1 for some complete setof coset representatives N ⊆ MIN. Further, if n < |A| then there is someN ′ ⊆ N with |N ′| = n and lub(N) = lub(N ′).
Theorem 1.7. MIN ⊆ Bx : x ∈ Nm0 , ||x||1 ≤ |A| − 1.
Corollary 1.8. If m = 1 then G = |A|B − A1 −B.
2 Adjacency
2.1 General
Definition 2.1. Let gi ∈ G be distinct vectors. A vector v ∈MR≥0is inside
the gi iff glb(gi) ≤ v ≤ lub(gi).
Definition 2.2. Let gi ∈ G be distinct vectors. A vector v ∈MR≥0is strictly
inside the gi iff glb(gi) ≺ v ≺ lub(gi).
Definition 2.3. The distinct vectors g1 . . . , gk ∈ G are adjacent iff there isno g ∈ G strictly inside the set gi. We write gi adjacent as ∧[gi]. Whenwe refer to only two vectors we may also write g1∧g2 for simplicity.
Definition 2.4. Let Bi = B ⊆ G | |B| = i and ∧[B]. If some B ∈ Bi ismaximal under inclusion we say it is a block. Let B be the set of blocks.
Definition 2.5. Let O(M) = lub(B) | B ∈ B. This is the O-set.
The following Theorem characterizes, based on n, how large blocks in Bcan be.
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Theorem 2.6. If Bi 6= ∅, then i ≤ n.
Proof. Suppose that i = n + 1. For B ∈ Bi let o = lub(B) such thatB = g1, . . . , gn+1 ⊆ G. Since each gi ∈ B contributes one or more Aek
components to o, by being maximal in that direction, we have by the pigeon-hole principle that one of the gi is strictly inside of B. Thus we have reacheda contradiction and Bn+1 = ∅. This argument holds when i > n + 1 as well.
Theorem 2.7. For all Bi ∈ Bi there exists B ∈ B such that Bi ⊆ B.
Proof. If Bi ∈ B, then we are done. Suppose that Bi /∈ B. Clearly Bi is notmaximal under inclusion; there must be some B′ ∈ Bi+1 such that Bi ⊆ B′.If again we have B′ /∈ B, the process continues. By Theorem 2.6 this processis finite and so ∃B′′ ∈ B such that Bi ⊆ B′′.
The following Corollary shows that for all g ∈ G there exists some B ∈ Bsuch that g ∈ B.
Corollary 2.8.⋃
B∈B B = G.
Proof. Clearly⋃
B∈B B ⊆ G. By definition B1 = G since ∀g ∈ G it is truethat ∧[g]. By Theorem 2.7 we then have G ⊆
⋃B∈B B so that
⋃B∈B B = G.
Theorem 2.9. Let B ⊆ G. Then lub(B) /∈MN0 if and only if ∧[B].
Proof. Suppose ∧[B]. Let L = A−1lub(B) and suppose that lub(B) ∈ MN0 .Since AL ∈ MN0 , we have that L ∈ A−1MN0 . Now let L′ = L − ε(In)1
where In is the identity matrix. Thus [≥ L] ⊆ A−1MN0 implies that L′ iscomplete in A−1MN0 and therefore AL′ is complete in MN0 . Since L′ < Lwe have AL′ < lub(B). Thus ∃g ∈ G(M) such that g is strictly inside B, acontradiction.
Now suppose that lub(B) /∈ MN0 . Then for all v ∈ [≺ lub(B)] we havev /∈ G(M) and therefore there are no frobenius vectors strictly inside B sothat ∧[B].
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Corollary 2.10. O(M) ∩MN0 = ∅.
Proof. By construction, for all o ∈ O(M) we have o = lub(B) where B ∈ B.Thus ∧[B] and the Corollary follows by Theorem 2.9.
2.2 Adjacency in two dimensions: the Ladder
The concept of adjacency can be used when looking at various properties ofthe sets of Frobenius vectors. For example, in two dimensions, it could beused to prove that since Frobenius vectors are incomparable to each other,they lie in a chain or a “ladder”, where the least upper bound of each con-secutive pair of vectors is the cut-off for each “step”.
The following two propositions have already been proved earlier in thepaper, and they are very useful in the two-dimensional case.
Proposition 2.11. Let g1, g2 ∈ G(M). Then g1∧g2 iff lub(g1, g2) /∈ (M)N0.
Proposition 2.12. O(M) ∩ (M)N0 = ∅.
Below follows a description of various properties of adjacent Frobeniusvectors in two dimensions and a proof that they all lie in a form of a “ladder”in two dimensions.
Definition 2.13. Let M = [a1, a2|B], for some vector g, and let the angleof g be the angle that g makes with a2. We write the angle of a as Θa.
Proposition 2.14. Let g1, g2 ∈ G(M). Then Θg1 = Θg2 iff g1 = g2.
Proof. If g1 = g2 then both g1 and g2 have the same coordinates, and there-fore the angles that they make with the origin would be equal.
Let Θg1 = Θg2 = α. Let Θa1 > Θa2 . Then, by the Definition 2.13,Θa1 ≥ Θg1 , ..., Θgk
≥ Θa2 , where k = |G(M)|. Both g1 and g2 lie on the sameline with the origin. Therefore let g2 = βg1, β > 1. Then let g1 = Σαiai,αi ≥ 0 since g1 ≥ 0. Then g2 = Σβiαiai. g2 − g1 = Σ(β − 1)αiai so g2 > g1
which is a contradiction, because unless the g-vectors are equal to each other,they can not be comparable.
Proposition 2.15. Let g1, g2 ∈ G(M), g1 6= g2. Let Θg1 > Θg2, a =lub(g1, g2) and b = glb(g1, g2). Then Θg1 > Θa > Θg2 and Θg1 > Θb > Θg2.
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Proof. Suppose A−1g1 = (x1, y1), A−1g2 = (x2, y2). Then Θg1 > Θg2 ⇔y1/x1 > y2/x2. Since g1||g2, A−1g1||A−1g2. Then y1 > y2 and x2 > x1.Since both a and b lie on the intersection of the cones of g1 and g2, bothA−1a and A−1b lie on the intersection of the cones of A−1g1 and A−1g2.A−1A = [[1, 0], [0, 1]] therefore the cones are parallel to the first quadrant,and A−1a = (x2, y1), A
Proposition 2.16. For any 3 vectors x, y, z, x 6= y 6= z, if Θx ≥ Θy ≥ Θz
then at least one of the following has to be true:1. y ≤ x2. y ≤ z3. y ≥ x4. y ≥ z5. glb(x, z) ≺ y ≺ lub(x, z)
Proof. Suppose that y||x and y||z, that is the first four conditions of theProposition 2.16 do not apply. Then let A−1x = [x1, y1], A−1z = [x2, y2], andA−1y = [x3, y3]. Since y||x then either:(1) x1 > x3 and y1 < y3 or(2) x1 < x3 and y1 > y3.Since y||z then either:(3) x2 > x3 and y2 < y3 or(4) x2 < x3 and y2 > y3.Note that since x1 < x2 and y1 < y2 only 3 out of the 4 combinationsare possible: (1) and (3), (2) and (3), (2) and (4). However the (1),(3)combination results in y3/x3 > y1/x1 ⇒ Θy > Θx which contradicts theassumption. The (2),(4) combination results in y3/x3 < y2/x2 ⇒ Θy < Θz
which also contradicts the initial assumption. Therefore, the only possiblerelationship is when x1 < x3 < x2 and y1 > y3 > y2. Therefore, sincelub(A−1x, A−1y) = [x2, y1] and glb(A−1x, A−1y) = [x1, y2], glb(x, z) ≺ y ≺lub(x, z).
Corollary 2.17. Let g1, g2 ∈ G(M), g1 6= g2. Then g1∧g2 iff there is noother g ∈ G(M), g 6= g1, g 6= g2 such that Θg1 > Θg > Θg2.
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Proof. Let g1∧g2. Since g1, g2, g ∈ G(M), g1||g2||g, and therefore by Propo-sition 2.16, glb(g1, g2) ≺ g ≺ lub(g1, g2)⇒ g is strictly inside g1 and g2 whichcontradicts the definition of adjacent vectors.
Let g1, g2 ∈ G(M) and let there be no g ∈ G(M) such that Θg1 >Θg > Θg2 . Then since by Proposition 2.15, Θg1 > Θlub(g1,g2) > Θg2 andΘg1 > Θglb(g1,g2) > Θg2 , there is no Frobenius vector that is strictly inside g1
and g2, therefore g1∧g2.
Theorem 2.18 (The Ladder). Let k = |G(M)|. Then all vectors gi ∈ G(M)can be relabeled in such a way that Θg1 > Θg2 > · · · > Θgk
and for each gi
and gi+1 under the new labeling system, gi∧gi+1.
Proof. By the Proposition 2.14 gi = gk iff Θgi= Θgk
⇒ if gi 6= gk thenΘgi6= Θgk. If Θgi
6= Θgk then either Θgi> Θgk
or Θgi< Θgk
. Therefore, sincethere are k different Frobenius vectors, there are k unique angles, which canbe listed by decreasing order of magnitude. Hence, g1, ..., gk can be relabeledin such a way that Θg1 > Θg2 > · · · > Θgk
. Then, under this labeling system,each pair gi and gi+1 has to be adjacent by the Proposition 2.17, since thereare no other vectors g ∈ G(M) for which Θgi
> Θg > Θgi+1.
Proposition 2.19. Let k = |G(M)|. Then |O(M)| = k − 1.
Proof. By the Theorem 2.18 all of the vectors gi ∈ G(M) can be relabeledin such a way that Θg1 > Θg2 > · · · > Θgk
⇒ there are k− 1 pairs of vectorsgi, gi+1 that are adjacent and correspond to a unique element of O. Anyother pair of vectors in G(M) would have at least one other vector with anangle that lies between their angles ⇒ by the Proposition 2.17 they wouldnot be adjacent ⇒ by the Proposition 2.12 their least upper bound wouldnot be in O(M). Therefore, |O(M)| = k − 1.
Proposition 2.20 (The Big Chain). Let n = |O(M)|. Then all of the pointsin Oi ∈ O(M) can be relabeled in such a way that ΘO1 > ΘO2 > · · · > ΘOn,and each pair of points Oi and Oi+1 lies on the cone of a unique Frobeniusvector.
Proof. By the Proposition 2.19, |G(M)| = n + 1, and by the Theorem 2.18all of the elements in G(M) can be relabeled in such a way that Θg1 >Θg2 > · · · > Θgn+1 . Therefore, there are n pairs of adjacent vectors of theform gi, gi+1, each corresponding to a specific element Oi such that, by theProposition 2.15, Θgi
> ΘOi> Θgi+1
. Therefore, all elements of O(M) can
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be relabeled in such a way that ΘO1 > ΘO2 > · · · > ΘOn . In fact, the chaincould be extended to Θg1 > ΘO1 > Θg2 > ΘO2 > · · · > ΘOi
> Θgi+1.
Proposition 2.21. If O1, O2 ∈ O(M), O1 6= O2 then O1||O2
Proof. Suppose that O1 is comparable to O2. WLOG let O1 > O2. But bythe definition of the least upper bound, for any ~v,~v > O2, ~v ∈M ⇒ O1 = O2
which is a contradiction to the hypothesis.
Applications of the Ladder and the Big Chain in the 4-2 case.
The fact that both the elements of G(M) and O(M) form a chain in twodimensions can be used to prove various properties of the g-set in two di-mensions. For example, in the 4 − 2 case, suppose that M1 = [a1, a2|b1, b2],and M2 = [a1, a2|b1, b2 + b1], and [a1, a2|b1] is dense. Then the chain prop-erty of the Frobenius vectors in two dimensions could be used to prove that|G(M2)| ≤ |G(M1)|.
Lemma 2.22. Let Oi ∈ O(M2) and Oi ∈ (M1)N0. Then Oi = d − b1 whered ∈ [a1, a2|b1 + b2]N0.
Proof. Let c? = lub(g1, g2), and let g1, g2 ∈ G(M2) such that g1∧g2. By theProposition 2.11, c? /∈ (M2)N0 . c? ∈ (M1)N0 . Let c? = c1a1+c2a2+c3b1+c4b2,where c1, c2, c3, c4 ∈ N0. There could be multiple ways to express c? interms of the vectors a1, a2, b1, b2 and let’s consider the one with the minimalcoefficient in front of b2. Suppose c3 > c4. Then c? = c1a1 + c2a2 + (c3 −c4)b1 + c4(b2 + b1) ⇒ c? ∈ (M2)N0 which contradicts the assumption thatc? /∈ (M2)N0 . Therefore, c3 < c4. Let c4 − c3 > 1. Let d = c? + b1.d = c1a1 + c2a2 +(c3 +1)b1 + c4b2. Since the coefficients in front of a1, a2 staythe same, c4 is still the least possible coefficient in front of b2. c3 + 1 < c4,therefore d can not be expressed as a multiple of a1, a2, b1, (b2 + b1) andtherefore d /∈ (M2)N0 . But d > c? which contradicts the hypothesis thatc? = lub(g1, g2) because d g for some g ∈ G(M2) ⇒ c3 = c4 − 1 andc? = c1a1 + c2a2 + (c4 − 1)b1 + c4b2 = c1a1 + c2a2 + c4(b1 + b2)− b1.
Lemma 2.23. For each Oi ∈ O(M2) there exists Oj ∈ O(M1) such thatOj = Oi − cb1; c ∈ N0.
Proof. By the Lemma 2.22, Oi = c1a1 + c2a2 + c3(b1 + b2) − b1. Let d? =Oi − c3b1 = c1a1 + c2a2 + c3b2 − b1. Oi /∈ (M2)N0 by the Proposition 2.12,
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therefore −1 is the maximum possible coefficient in front of b1 for Oi ⇒ −1is the maximum coefficient in front of b1 for d?, hence d? /∈ (M1)N0 , since oneof the coefficients is negative. Now follows a proof by induction that everyvector greater than d? is in (M1)N0 .Let c? = A−1Oi and c′ = A−1d?. Let a1 = A−1a1 = [1, 0], a2 = A−1a2 =[0, 1], b1 = A−1b1, b2 = A−1b2.Base case: c′ + [ε, 0] and c′ + [0, ε] ∈ A−1(M1)N0 .Since [a1, a2|b1] is dense, then A−1[a1, a2|b1] is also dense ⇒ [ε, 0] = n1a1 +n2a2 +n3b1; n1, n2, n3 ∈ Z. c? + [ε, 0] ∈ A−1(M2)N0 . c? + [ε, 0] = c1a1 + c2a2 +(c4−1)b1+c4b2+n1a1+n2a2+n3b1 ⇔ (c1+n1)a1+(c2+n2)a2+(c4−1+n3)b1+c4b2 ∈ A−1(M2)N0 ⇒ c1+n1 ≥ 0; c2+n2 ≥ 0; and c4−1+n3 ≥ c4 ⇒ n3−1 ≥ 0.Now c′ + [ε, 0] or c′ + [0, ε] would equal to (c1 + n1)a1 + (c2 + n2)a2 + (n3 −1)b1 + c4b2, therefore, since all of the coefficients are non-negative, c′ + [0, ε]and c′ + [ε, 0] ∈ A−1(M1)N0 .Assume that c′+[a, b] ∈ A−1(M1)N0 , where [a, b] > 0. c?+[a, b] ∈ A−1(M2)N0 .Let c′ + [a, b] = d1a1 + d2a2 + d3b1 + d4b2, where d1, d2, d3, d4 ∈ N0. c? +[a, b] = d1a1 + d2a2 + (d3 + c4)b1 + d4b2. c? + [a, b] + [0, ε] ∈ A−1(M2)N0 .Both c? + [a, b] and c? + [a, b] + [0, ε] ∈ A−1(M1)Z ⇒ [0, ε] ∈ A−1(M1)Z ⇒[0, ε] = s1a1 + s2a2 + s3b1 + s4b2 ⇒ s1, s2, s3, s4 ∈ Z. By the base casec′ + [0, ε] ∈ A−1(M1)N0 ⇒ (c1 + s1)a1 + (c2 + s2)a2 + (s3− 1)b1 + (s4 + c4)b2 ∈A−1(M1)N0 ⇒ s3−1 ≥ 0⇒ s3 ≥ 1. c?+[a, b]+[0, ε] = (d1+s1)a1+(d2+s2)a2+(d3 +c4 +s3)b1 +(d4 +c4)b2 ∈ A−1(M2)N0 ⇒ d1 +s1 > 0; d2 +s2 > 0; d4 +c4 >0⇒ c′ + [a, b] + [0, ε] = (d1 + s1)a1 + (d2 + s2)a2 + (d3 + s3)b1 + (d4 + s4)b2.Since d3 ≥ 0 and s3 ≥ 1 ⇒ d3 + s3 ≥ 1 ⇒ c′ + [a, b] + [0, ε] ∈ A−1(M1)N0 .The argument for c′ + [a, b] + [ε, 0] is exactly the same. Since c′ /∈ A−1(M1)N0
and c′ is complete, c′ ∈ O(A−1M1)⇒ d? ∈ O(M1).
Lemma 2.24. Let Oj1 , Oj2 ∈ O(M1) such that Oj1 +c1b1 = Oi1 , Oi1 ∈ O(M2)and Oj2 + c2b1 = Oi2 , Oi2 ∈ O(M2), where c1, c2 ∈ N0. Then if Oj1 = Oj2
then Oi1 = Oi2.
Proof. Let Oj1 = Oj2 = p. Then Oi1 = p + c1b1 Oi2 = p + c2b1. SupposeOi1 6= Oi2 then c1 6= c2. WLOG let c2 > c1. Then Oi2 = Oi1 + (c2 − c1)b1 ⇒Oi2 > Oi1 but that contradicts Proposition 2.21, because Oi1 can not becomparable to Oi2 , therefore Oi1 = Oi2 .
Proposition 2.25. |G(M2)| ≤ |G(M1)|.
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Proof. Let |O(M2)| = k. Since, by Lemma 2.23 there is a function f :O(M2)→ O(M1) And by Lemma 2.24, f is injective ⇒ |O(M1)| ≥ |O(M2)|.Since by the Proposition 2.19, |G(M2)| = |O(M2)| + 1 and |G(M1)| =|O(M1)|+ 1, |G(M2)| ≤ |G(M1)|.
Theorem 2.26. Let M1 = [a1, a2|b1, b2] and M2 = [a1, a2|b1, b2 +c ·b1], wherec ∈ N0, and [a1, a2|b1] is dense. Then |G(M2)| ≤ |G(M1)|.
The last theorem shows that the number of Frobenius vectors can onlydecrease as one of the b-vectors is added to another in the 4− 2 case by themeans of establishing a relationship between the elements of the O-sets forthe two sets of generating vectors. The concept of the chain could possibly beused for further research about the properties of the g-set in two dimensions,and the concept of adjacency can be used for the multi-dimensional proofs.
3 Order of bi and the MIN set
Much of the time, information regarding frobenius vectors comes directlyfrom a knowledge of the MIN set. The relationship between frobenius vec-tors and the least upper bounds of subsets of MIN is fundamentally im-portant in this process. From some of the basic definitions of MIN we canprove theorems concerning the calculation of these vectors, specifically incases where a unique frobenius vector is present. The following argumentsare based on the relationship of the orders of the B vectors to the set MIN .
Consider the monoid MN0 where M = [A|B]. Let bi be the columns of B.We say ordA (bi) is the smallest k ∈ N such that kbi ≡ 0 mod A. We definethe following function:
Ω(M) =m∏
i=1
ordA (bi) .
Let SM = ∑m
i=1 cibi | 0 ≤ ci < ordA (bi). We see that |SM | = Ω(M). Aknown theorem states that the set MIN is totally dependent on B. In fact,we show that MIN is a subset of a specific finite subset of BN0 , namely SM .
Proposition 3.1. MIN ⊆ SM .
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Proof. By Theorem 1.7 we have that MIN ⊆ BN0 so now suppose thatv ∈ MIN but v /∈ SM . Since v ∈ MIN ⊆ BN0 there exists some v′ ∈ Nm
0
such that v = Bv′. Now for some j ∈ [1, n], it must be that (v′)j ≥ ordA (bj).Therefore v ≡> v − bjordA (bj) and thus v /∈ MIN , a contradiction. HenceMIN ⊆ SM .
Theorem 3.2. If Ω(M) < |A|, then G(M) = ∅. Furthermore, if gcd(M) = 1and Ω(M) = |A|, then G(M) =
∑mi=1(ordA (bi)− 1)bi − A1.
Proof. Suppose that Ω(M) < |A|. Thus no subset of SM can be a completeset of minimal residues, and therefore there must exist some equivalence classω that has no representatives in the monoid. Thus there can be no completepoints and hence G(M) = ∅.
Now, let Ω(M) = |A|. Due to gcd(M) = 1 we know that G(M) isnonempty and therefore there must be at least one minimal complete setof residues so that |MIN | ≥ |A|. By Proposition 3.1, MIN ⊆ SM , and|SM | = Ω(M) we have that
|A| ≤ |MIN | ≤ |SM | = Ω(M) = |A|
So |MIN | = |A| and particularly, MIN = SM . Thus for every g ∈ G(M),there is L ⊆ MIN with |L| = |A| such that g + A1 = lub(L), but thenL = MIN so that
g + A1 = lub(MIN)⇒ g =m∑
i=1
(ordA (bi)− 1)bi − A1
Hence we have G(M) = ∑m
i=1(ordA (bi)− 1)bi − A1.
Several special cases involving a diagonal A matrix follow rather nicelyfrom the above theorem.
12
Corollary 3.3. Let A be a in Smith-Normal form with diagonal [k1, . . . , kn]where ki ∈ N0. Let B be the n × n matrix (bij) with columns bj. Also, let
kp
kn−q+1| bpq for n + 1 < p + q ≤ 2n, and suppose gcd(M) = 1; then there is a
unique frobenius vector
G(M) =
(n∑
j=1
(kj − 1)bn−j+1
)− A1
.
Proof. Since knbi1 ≡ 0 mod kn for all i, and all other diagonal entries dividekn, we have that ordA (b1) ≤ kn. For all j such that 1 ≤ j < n we will showthat ordA (bn−j+1) ≤ kj.
For some j such that 1 ≤ j < n we choose some column bn−j+1; considerkjbn−j+1 mod kj. Clearly for the entries bi(n−j+1) where 1 ≤ i ≤ j we havethat kjbi(n−j+1) ≡ 0 mod ki since all such ki | kj. Now when j < i ≤ n we
have n + 1 < i + (n− j + 1) ≤ 2n so that the condition kp
kn−q+1| bpq applies.
Thus there exists y ∈ N such that
kjbi(n−j+1) = kj
(ki
kn−(n−j+1)+1
)y
= kj
(ki
kj
)y
= kiy ≡ 0 mod ki
Therefore, for all i we have ordki
(bi(n−j+1)
)≤ ki ⇒ ordA (bn−j+1) ≤ kj for
all 1 ≤ j ≤ n. Hence for all j we have ordA (bn−j+1) ≤ kj. This shows that
Ω(M) ≤n∏
j=1
kj = |A|
But since gcd(M) = 1, we have |A| ≤ |MIN | ≤ Ω(M) ≤ |A| so that Ω(M) =
|A| and Theorem 3.2 shows that G(M) =(∑n
j=1(kj − 1)bn−j+1
)− A1
.
Corollary 3.4. Let A = dI where d ∈ N and I the identity matrix, B thesame dimension as A, and suppose gcd(M) = 1; then G(M) = (d− 1)B1−A1.
13
Proof. Follows directly from Corollary 3.3.
Later sections will discuss the cases when there exists some s ∈ N suchthat sb1 ≡ b2 mod A. Here we explore the case when no such s exists. Inthe next section, we see how any monoid can be reduced into a case wheresuch an s exists, but under certain circumstances the following provides asimplified solution for cases in which there is a unique frobenius vector.
Theorem 3.5. Without loss of generality, let 1 < ordA (bj) ≤ ordA (bi). Ifthere is an si ∈ N such that sibi ≡ bj mod A for i, j ∈ [1, m] with i 6= j thenfor every pair bi, bj we have gcd(ordA (bi) , ordA (bj)) = ordA (bj).
Proof. Now suppose that there exists an si ∈ N such that sibi ≡ bj mod A.Then we have
ordA (bj) = ordA (sibi) =ordA (bi)
gcd(si, ordA (bi))
Thus ordA (bj) | ordA (bi) so that gcd(ordA (bi) , ordA (bj)) = ordA (bj).
Corollary 3.6. Without loss of generality, let 1 < ordA (bj) ≤ ordA (bi). Ifall bi, bj where i, j ∈ [1, m] with i 6= j are pairwise relatively prime, then thereis no si ∈ N such that sibi ≡ bj mod A.
Proof. The proof follows immediately from Theorem 3.5.
We now demonstrate a partial solution for the case when there is no suchsi ∈ N so that sibi ≡ bj mod A for i, j ∈ [1, m] with i 6= j, namely when theorders of the bi are pairwise relatively prime.
Theorem 3.7. A monoid MN0 such that ordA (bi) are pairwise relativelyprime will have at most one frobenius vector, G(M) = lub(SM)− A1.
14
Proof. Consider ordA (bi). We know that for all i, ordA (bi) | |A|. Suppose|A| =
∏ri=1 pki
i where the pi are distinct primes. So all possible orders forthe bi must be some combination of elements from the multiset of primedivisors of |A|. Furthermore, when ordA (bi) are pairwise relatively prime,the orders of the bi can have no terms in common. Thus Ω(M) ≤ |A|. Whengcd(M) 6= 1 some equivalence class has no representatives in MN0 and soG(M) = ∅. If gcd(M) = 1, by Theorem 3.2 we have G(M) = ∅ whenΩ(M) < |A| and G(M) = lub(SM)− A1 when Ω(M) = |A|.
4 Useful Theorems
The following Theorems are general, basic results which appear in the proofsof later concepts.
Let N ⊂MIN be a complete set of residues. We’ve seen in Theorem 1.6that for any g ∈ G, we can write g + A1 = lub(N) for some complete set ofresidues. We will now prove two other properties of lub(N).
Theorem 4.1. lub(N)− A1 is complete.
Proof. Let v = lub(N) − A1, and let u ∈ [ v]. Let w be the element in Ncongruent to u. We know that u v, v + A1 ≥ w and u ≡ w. Thus foreach i = 1, . . . , n we have (u)i > (v)i, (v)i + 1 ≥ (w)i, and (u)i − (w)i ∈ Z.Now (u)i − (w)i ≥ (u)i − ((v)i + 1) > −1, and (u)i − ((v)i + 1) ≥ 0. Thusw ≤ v + A1 ≤ u and u ∈MN0 . Finally, [v] ⊂MN0 and v is complete.
Theorem 4.2. Suppose ω = lub(N), where ω ∈ N . Then ω − A1 ∈ G.
Proof. By Theorem 4.1, ω −A1 is complete. Thus there exists a g ∈ G suchthat g ≤ ω − A1. By Theorem 1.6, there exists a complete set of residuesN
′ ⊂ MIN such that g = lub(N′)− A1. Choose ω
′ ∈ N′such that ω ≡ ω
′.
But ω′ ≤ lub(N
′) = g + A1 ≤ ω so ω = ω
′. Now lub(N
′) ≥ ω
′= ω. Thus
ω − A1 = lub(N′)− A1 ∈ G.
15
5 Reduction formula
When solving the Frobenius number problem with three generators, Selmerfound it useful to consider only the case where the three generators are rel-atively prime. We would like to take a similar approach to the Frobeniusvector problem where m = n = 2.
In 1960, Johnson proved that f(a1, da2, . . . , dak) = d·f(a1, . . . , ak), wherea1 need not be the smallest generator. Using this reduction formula, theFrobenius problem can be reduced to the case where any k − 1 of the kgenerators have a gcd of 1. When k = 3, this reduces the problem down tothe case were the three generators are pairwise relatively prime. This is thereduction formula Selmer used to solve the three generator case.
The next two theorems provide a perfect generalization of Johnson’s for-mula into multiple dimensions. Loosely speaking, the first theorem considersthe case where all vectors but a column of A are multiples of matrix D, andthe second theorem considers the case where all vectors but a column of Bare multiples of D. For notation simplicity we assume without loss of gen-erality that the column receiving special treatment is either the first columnof M in the first theorem, or the last column of M in the second theorem.
Theorem 5.1. Let M = [a1|M ′] and N =[
Da1
|D|
∣∣∣DM ′]
be integer matrices
where D ∈ Mn[Z] with |D| 6= 0 and gcd(N) = 1 with M or N simplicial.Then D · F (M) = F (N).
To prove this theorem we will first prove that the transformations betweenS(M)N0 and S(N)N0 preserve inequality, congruence, least upper bounds, theMIN set, the G-set, and finally the F-set. Let A and A
′be the left-most
n by n submatrices of M and N respectively. Because in general M and Nhave different left n by n sub-matrices, the symbols ≥, ≡, lub, and MIN areall ambiguous. For the duration of this proof we will always use a subscriptto specify the context of each.
Lemma 5.2. For any vectors v and w, we have v ≥M w ⇐⇒ Dv ≥N Dw.
Proof. Notice that for any vector u, we know that: u ≥M 0⇐⇒ There exist non-negative reals α1, . . . , αn such that
∑αiai = u
⇐⇒ There exist non-negative reals α1, . . . , αn such that∑
αiDai = Du⇐⇒ Du ≥N 0.Now by letting u = v−w we can see that v−w ≥M 0 ⇐⇒ D(v−w) ≥N 0,which proves the lemma.
16
All columns of M are ≥M 0 ⇐⇒ all columns of N are ≥N 0. Becauseone of M and N is simplicial, they must both be simplicial.
Lemma 5.3. For any vectors v and w, we have v ≡M w ⇐⇒ Dv ≡N Dw.
Proof. Notice that for any vector u, we know that: u ≡M 0=⇒ There exist integers c1, . . . , cn such that
∑ciai = u
=⇒ There exist integers c1, . . . , cn such that∑
ciDai = Du=⇒ Du ≡N 0.Now by letting u = v−w we can see that v−w ≡M 0 =⇒ D(v−w) ≡N 0,which proves that multiplication by D preserves congruence. Because S(N)N0
is dense, it contains all residue classes mod A′, as does the set D · BN0 .
Thus multiplication by D is onto. Finally, |A′| =∣∣∣Da1
|D| , Da2, . . . , Dam
∣∣∣ =
1|D|
∣∣∣Da1, Da2, . . . , Dam
∣∣∣ = 1|D| |DA| = |A|, so the number of residues mod A
is equal to the number of residues mod A′, and thus multiplication by D is
a bijection of residue classes.
Lemma 5.4. For any vectors v1, . . . , vk, we have D · lubM(v1, . . . , vk) =lubN(Dv1, . . . , Dvk).
Proof. For i = 1, . . . , k we have lubM(v1, . . . , vk) ≥M vi, so by Lemma 5.2 D ·lubM(v1, . . . , vk) ≥N Dvi, thus D · lubM(v1, . . . , vk) ≥N lubN(Dv1, . . . , Dvk).Similarly, for i = 1, . . . , k we have lubN(Dv1, . . . , Dvk) ≥N Dvi, so by Lemma5.2 D−1 · lubN(Dv1, . . . , Dvk) ≥M vi, and D−1 · lubN(Dv1, . . . , Dvk) ≥M
Proof. It has been proven that both sets are contained in D · BN0 . The setsare the elements that are minimal in their congruence class. From Lemmas5.2 and 5.3 the two sets both use the same congruence condition and thesame partial ordering, and thus the two sets are equal.
We now have proven enough structural similarities to prove Theorem 5.1.
17
Proof. Let T be the set of points that can be written as the lubM of a completeset of residues from MINM , and let T
′be the set of points that can be written
as the lubN of a complete set of residues from MINN . It has been proventhat G(M)+A1 is contained in T . Also, all vectors in T are g-complete, thusG(M) + A1 is equal to the set of minimal elements in T . By Lemma 5.2 thisimplies that D ·G(M)+DA1 is equal to the set of minimal elements in D ·T .Similarly, G(N) + A
′1 is equal to the set of minimal elements in T
′. From
Lemmas 5.4 and 5.5, we can see that D · T = T′. Thus D ·G(M) + DA1 =
G(N) + A′1. This result can be stated more succinctly in terms of the F -
set. It becomes D · (F (M)− A1 −B1) + DA1 = F (N)− A′1 −DB1 + A
′1 or
D · F (M) = F (N).
Because the result of the second theorem mirrors the result of the firsttheorem, we will reuse much of the notation. We certainly are not assumingthat Lemmas 5.2, 5.3, 5.4 and 5.5 remain valid in this new context.
Theorem 5.6. Let M = [M ′, bm] and N =[DM ′
∣∣∣Dbm
|D|
]be integer matrices
where D ∈ Mn[Z] with |D| 6= 0 and gcd(N) = 1 with M or N simplicial.Then D · F (M) = F (N).
First we will see that multiplication by D preserves both types of inequal-ity.
Lemma 5.7. For any vectors v and w, we have v ≥M w ⇐⇒ Dv ≥N Dw,and v M w ⇐⇒ Dv N Dw.
Proof. Notice that for any vector u, we know that: u ≥M 0⇐⇒ There exist non-negative reals α1, . . . , αn such that
∑αiai = u
⇐⇒ There exist non-negative reals α1, . . . , αn such that∑
αiDai = Du⇐⇒ Du ≥N 0.Now by letting u = v−w we can see that v−w ≥M 0 ⇐⇒ D(v−w) ≥N 0,which proves the first part of the lemma. The exact same argument withpositive real for the αi proves the second part.
All columns of M are ≥M 0 ⇐⇒ all columns of N are ≥N 0. Becauseone of M and N is simplicial, they must both be simplicial. Now we willprove Theorem 5.6.
18
Proof. Throughout this proof we will consider congruence (mod D), where wedefine congruence as a ≡D b iff a− b ∈ D ·Zn. We are considering the cosetsof Zn
D·Zn under addition, so we have an equivalence relation and a group underaddition. Suppose for contradiction that SN does not contain some residues(mod D), say r. Now the sets SN and r+DZ are disjoint. Let v be any vectorin Rn. We know that both DA1 ≡D 0 and DA1 = Da1+· · ·+Dan N 0. Thusfor sufficiently large k ∈ Z, we have v− r ≺N k ·DA1, or v ≺N k ·DA1 + r ∈Zn/SN . Thus v is not complete and |G(N)| = 0. This contradicts the factthat gcd(N) = 1. Thus SN contains all residues (mod D). All but the lastcolumn of N is ≡D 0, thus Dbm
|D| is a generator for all |D| congruence classes.
We will define the vector function f(x) = (|D| − 1) Dbm
|D| + D · x, which
has inverse f−1(x) = D−1(x− (|D| − 1) Dbm
|D|
). By Lemma 5.7 this function
preserves inequality and in both directions.First we will see that if g is complete in MN, then f(g) is complete in
NN. Let v′
be an integer vector such that v′ N f(g). For some k =
0, . . . , |D|−1 we have v′ ≡D kDbm
|D| . We know that v′−kDbm
|D| N f(g)−kDbm
|D| ≥f(g) − (|D| − 1) Dbm
|D| = D · g, and that v′ − kDbm
|D| ≡D 0. Thus we can write
v′ − kDbm
|D| = D · v where v is an integer vector with v M g. Because g iscomplete in MN, v ∈MN and thus D · v ∈ D ·MN ⊂ NN. Finally, by writingv′= D · v + kDbm
|D| we see that v′ ∈ NN and f(g) is complete.
Conversely, we will see that if f(g) is complete in NN, then g is complete inMN. Let v be an integer vector such that v g. Now f(v) f(g). Becausef(g) is complete, there exist non-negative integers c1, . . . , cn+m such that
f(v) =n∑
i=1
ciDai +m−1∑i=1
ci+nDbi + cn+mDbm
|D| . Because f(v) ≡D (|D| − 1) Dbm
|D| ,
we have cn+m ≡ (|D| − 1) (mod |D|). Letting c′n+m = cn+m−|D|+1
|D| ∈ N we
have D · v = f(v)− (|D| − 1) Dbm
|D| =n∑
i=1
ciDai +m−1∑i=1
ci+nDbi + c′n+mDbm and
thus v ∈MN and g is complete.We will show that f(G(M)) = G(N) by showing that the sets are con-
tained in one another.Let g ∈ G(M). Because f(g) is complete in NN, there exists some g
′such
that f(g′) ≤N f(g) and f(g
′) ∈ G(N). Now g
′is complete in MN, so there
exists some g′′ ∈ G(M) such that g
′′ ≤M g′ ≤M g. Now g
′′, g ∈ G(M), so
g′′
= g′= g, and f(g) = f(g
′) ∈ G(N).
Now let f(g) ∈ G(N). Because g is complete in MN there exists some
19
g′ ∈ G(M) such that g
′ ≤ g. We’ve shown that f(g′) ∈ G(N) so f(g
′) ≤ f(g)
implies that f(g′) = f(g) and g = g
′ ∈ G(M). Thus f(G(M)) = G(N).Writing out function f we have (|D| − 1) Dbm
|D| + D ·G(M) = G(N). Sub-
stituting in the formula for the F-set we get (|D| − 1) Dbm
|D| + D · (F (M) −M
′1 − bm) = F (N)−DM
′1 − Dbm
|D| or D · F (M) = F (N).
Theorems 5.1 and 5.6 are equivalent to the following unification:
Theorem 5.8. Let M1, M2 be matrices with n rows and let c be a n × 1
vector. Let M =[M1
∣∣c∣∣M2
]and N =
[DM1
∣∣Dc|D|
∣∣DM2
]be n × (n + m)
integer matrices where D ∈ Mn[Z] with |D| 6= 0 and gcd(N) = 1 with M orN simplicial. Then D · F (M) = F (N).
Proof. When c is one of the first n columns of M the result follows fromTheorem 5.1. When c is one of the last m columns of M the result followsfrom 5.6.
Next, we will find what the transformation from N to M does to the gcd’sof submatrices of M .
Theorem 5.9. Define M , N , c, and D as in Theorem 5.8. Let M′and N
′
be sub-matrices of M and N respectively which are taken from correspondingcolumns of M and N . If column c is not contained in M
′, we have gcd(M
′) =
gcd(N′)
|D| . Otherwise, gcd(M′) ≤ gcd(N
′).
Proof. First, we will show that the determinants of all n × n submatriceswhich do not include c have been reduced by a factor of |D|, while thedeterminants of all other n × n submatrices have not been changed. Con-sider an n × n submatrix of M which does not contain c. It is D times itscorresponding matrix in M , and thus the first determinate is |D| times as
large. Next, consider a n × n submatrix of N containing Dc|D| , say
[Dc|D|
∣∣DS].
Its determinate is the same as that of its corresponding submatrix in M :∣∣∣ [Dc|D|
∣∣DS] ∣∣∣ = |D|
∣∣∣ [ c|D|
∣∣S] ∣∣∣ =∣∣∣ [c|S]
∣∣∣.Now suppose that M
′does not contain c. Here, DM
′= N
′, so by Lemma
23 of last summer’s paper, |D| gcd(M′) = gcd(DM
′) = gcd(N
′). Finally,
suppose M′contains c. Consider all pairs of corresponding n×n submatrices
20
of M′and N
′. Either the determinants are equal or the determinant of the
submatrix of N′
is |D| times as large. In all cases, the first determinantdivides the second, thus gcd(M
′) ≤ gcd(N
′).
What remains is the show how Theorem 5.8 can be used to reduce theFrobenius vector problem into the case where any n + m − 1 of the n + mgenerating vectors have a gcd of 1.
Theorem 5.10. Let N ∈ Mn×(n+m)[Z] with gcd(M) = 1. Then there existsD ∈Mn[Z] and M ∈Mn×(n+m)[Z] where every n× (n + m− 1) submatrix ofM has a gcd of 1 and F (N) = D · F (M).
Proof. Suppose that N =[a′1
∣∣N ′]with gcd(N
′) = d. Consider the matrix[
da′1,∣∣N ′]
. Every minor of this matrix which is also a minor of N′is divisible
by d by assumption, as are all other minors of N because they contain thecolumn da
′1. Because gcd(N
′) = d, we must have gcd
(da
′1
∣∣N ′)= d. By
Theorem 4 from last summer’s paper, there exist some D ∈ Mn(Z) with|D| = d and some M =
[a1
∣∣M ′] ∈ Mn,n+m(Z) with a gcd of d|D| = 1, such
that[da
′1
∣∣N ′]= D ·
[a1
∣∣M ′]=[Da1
∣∣DM′]
. The Frobenius vector problem
is now reduced in the following manner: F(a′1
∣∣N ′)= F
(Da1
|D|
∣∣DM′)
= D ·F(a1
∣∣M ′), where the first equality is because the matrices have been chosen
to be equal and the second equality is by Theorem 5.8.
By Theorem 5.9, 1 = gcd(N′)
|D| = gcd(M′). Also by Theorem 5.9, we
have not increased any of the other gcd’s. We perform the same processwith every other set of n + m − 1 vectors. We will be left with F (N) =D1D2 · · ·Dn+m ·F (M) where M is a matrix such that every n× (n + m− 1)submatrix has a gcd of 1.
Notice that Theorem 5.8 performs a linear transformation on the vectorsfollowed by multiplying one vector by a constant. When used repeatedly thisprocess performs a linear transformation on the directions of the vectors, butnot on the magnitudes of the vectors.
6 Small General Results
Theorem 6.1. Let g0 be the unique frobenius vector of M0N such that g0 +
A1 = lubwi. Let M = [M0b2]. If for any k, (b2)k > (wi)k, then g0 ∈ G(M)
21
Proof: Let g0 be such that g0 + A1 = lub(wi). If b2 ∈ M0N, be an earlier
theorem, M0N = MN, so G(M0) = G(M). Let b2 6= M0
N. For any coordinatek, without loss of generality, (b2)k > (wi)k.Then for any m ∈ MIN , m = (c1b1 + c2b2)k > (wi)k for c2 > 0 because(c1b1 + c2b2)k > (b2)k > (wi)k. So lub(wi) is still minimal ⇒ g0 ∈ G. .
Lemma 6.2. Set M ′ = [M |b1]. If b1 ∈ MN then MN = M ′N and G (M) =
G (M ′).
Proof: Let the ai’s denote the n + m columns of M . Suppose b1 ∈ MN.
Then we have that b1 =n∑
j=1
cjaj for some non negative integers cj. Consider
the monoid generated by M ′ = Mb1. Every element bk in the monoid of M ′,
S ′, can be expressed as bk =n∑
i=1
diaj + dkb1 where dk and the di’s are also
non negative integers. But then, for bk ∈ S ′,
bk =n∑
i=1
diaj + dkb1
=n∑
i=1
diai + dk
n∑j=1
cjaj
=n∑
j′=1
c′j′aj′ .
But then for any bk ∈ S ′, bk ∈ S. Every bk in S is in S ′ because the columnsof M are contained in the columns of M ′. So, S = S ′. Therefore, these twoidentical monoids have identical Frobenius numbers, so G (M) = G (M ′). .
Lemma 6.3. Suppose that M = [AB] and M ′ = [ABbi] with bi /∈ MIN ′.Then MN = M ′
N.
Proof: Suppose bi /∈ MIN ′. Then there exists some v with v ≡ bi andv < bi. v < bi and v ∈MIN ′ ⇒ v =
∑cjbj with j 6= i. bi = v+d1a1+d2b2 =∑
cjbj + d1a1 + d2a2 ∈MN. Therefor bi ∈MN. By lemma 6.2, MN = M ′N.
Lemma 6.4. If x ∈ MIN , with ω ∈ MN, then for some u = (x− ω) ∩MN,u ∈MIN .
22
Proof: It is equivalent to show that if u 6∈ MIN , then x 6∈ MIN .Suppose u 6∈ MIN , then there is some w ∈ MN with w ≡ u mod A andw < u. But then w +ω ≡ u+ω mod A and w +ω < u+ω. So x = u+ω 6∈MIN .
Lemma 6.5. Let M = [Ab1b2] and M ′ = [Ab1(b1 + b2)]. If w ∈ MIN andw ∈M ′
N, then w ∈MIN ′.
Proof: Suppose towards contradiction that w ∈ MIN and w ∈ M ′N,
but w /∈ MIN ′. So there is some u ∈ M ′N, with u ≡ w and u < w. u =
c1a1 + c2a2 + d1b1 + d2(b1 + b2) = c1a1 + c2a2 + (d1 + d2)b1 + d2b2 ∈ MN.But then u ∈ MN with u ≡ w and u < w ⇒ w /∈ MIN . But this is acontradiction of w ∈MIN .
Lemma 6.6. Let M = [Ab1b2]. Let g0 be the unique frobenius vector gen-erated by the submonoid M0 = [Ab1]. Suppose b2 /∈ M0
N, with (b2)x > (b1)x
and b2 ≤ lub(jbi|1 ≤ j ≤ |A|). Then there exists some g1 ∈ G(M) with(g1)y < (g0)y.
Proof: If b2 /∈ M0N ⇒ b2 ≡ kb1 with 1 ≤ k ≤ |A|. If b2 /∈ MIN , then
b2 = c1a1 + c2a2 + d1b1 ⇒ b2 ∈ M0N, which contradicts our hypothesis. So
b2 ∈MIN .
So b2 ≡ kb1 and b2 incomparable to kb1. Likewise, b2 ≡ kb1 ⇒ b2 + b1 ≡(k + 1)b1. Likewise, b2 is incomparable to kb1 ⇒ b2 + b1 is incomparable to(k + 1)b1. Therefor, b2 + b1 ∈MIN for k < a1,1a2,2.
Likewise, for all b2 + d1b1 such that 0 ≤ d1 ≤ (a1,1a2,2 − k), b2 + d1b1 ≡(k + d1)b1 and b2 + d1b1 incomparable to (k + d1)b1, so b2 + d1b1 ∈MIN .
So there is subset of MIN , say W such that W = b1, 2b1, . . . , (k −1)b1, b2, b2+b1, b+2+2b1, . . . , b2+(|A|−k)b1. By Theorem 4.1, lub(W )−A1 =h is g-complete.
However (lub(W ))y < (lub(jb1))y because (lub(W ))y = ((k − 1)b1)y or(lub(W ))y = ((b2 +(|A|−k)b1)y. In either case, (lub(W ))y < (lub(jb1))y =(|A|b1)y.
23
So there is some h = lub(W ) − A1, with h g complete by Theorem 4.1.Either h is a g vector or there is some gk < h with gk < h. Either way thereexists some g1 ∈ G(M) with (g1)y < (g0)y.
Lemma 6.7. Let M = [Ab1b2] with diagonal A. Let g0 be the unique frobeniusvector generated by the submonoid M0 = [Ab1]. Suppose b2 /∈ M0
N, with(b2)y > (b1)y and b2 ≤ lub(jbi|1 ≤ j ≤ |A|). Then there exists someg1 ∈ G(M) with (g1)x < (g0)x.
Proof: This follows from the argument above, exchanging x and y.
Lemma 6.8. Let M = [Ab1b2] with diagonal A. Let g0 be the unique frobeniusvector generated by the submonoid M0 = [Ab1]. If b2 /∈M0
N and |G(M)| = 1,then g0 /∈ G(M).
Proof: By lemmas 6.6 and 6.7, there is some g1 ∈ G(M), and wlog,with (g1)x < (g0)x. If (g1)y > (g0)y, then g0 and g1 are incomparable, so|G(M)| 6= 1. So (g1)y ≤ (g0)y. But then g1 < g0, so g0 /∈ G(M).
Theorem 6.9. Let g0 ∈ G(M) with M = [AB] and let M ′ = [ABb1]. Ifb1 g0 + A1, then g0 ∈ G(M ′).
Proof: If b1 > g0 + A1, then b1 ∈ MN. By lemma 6.2, MN = M ′N, so
G(M) = G(M ′); hence, g0 ∈ G(M ′).
If b1 > g0 + A1 and b1 g0 + A1, then b1 is incomparable to g0. Withoutloss of generality, assume (g0)x < (b1)x and (g0)y > (b1)y. By theorem 6.1,g0 ∈ G(M ′).
7 A further reduction in the case where b2
lies along an a vector
Theorem 7.1. Suppose m = n = 2 and b1 = ka2 for some k ∈ R+. Then
there exist D ∈M2[Z] and a, b, d ∈ N0 such that F (M) = D ·[
1 0 0 10 a b d
]By Theorem 5.10, there exist D1 ∈ Mn[Z] and M
′ ∈ Mn×(n+m)[Z] whereevery 2×3 submatrix of M has a gcd of 1 and F (M) = D1 ·F (M
′). Let M
′=[
a′1, a
′2, b
′1, b
′2
]. We still have b
′1 = k
′a′1 for some k
′ ∈ R because Theorem 5.10performs a linear transformation on the directions of the generators.
24
Because a′1 and b
′1 are integer vectors, k
′= b
afor some relatively prime
a, b ∈ N. Using the Euclidean Algorithm, we can write x = 1aa′2 as a linear
combination of a′2 and b
′1, thus 1
aa′2 is an integer vector. Next, we see that
a′2 = a · x and b
′1 = b · x. Let D2 =
[a′1, x]. We have gcd
[a′1, a
′2, b
′1
]= 1
and[a′1, a
′2, b
′1
]N0⊂ D2,N0 , thus D2,N0 is dense, |D2| = 1, and D2 is in-
vertible over Z. Let D−12 b
′2 =
[cd
]. By Theorem 5.8, we have F (M
′) =
F
[D2 ·
[1 0 00 a b
] ∣∣∣ b′2
|D2|
]= D2 · F
[1 0 0 c0 a b d
].
However, gcd(a′2, b
′1, b
′2) = 1, and the previous application of Theorem 5.8
does not increase this gcd by Theorem 5.9. Thus gcd
[0 0 ca b d
]= 1, we
must have c = 1. From our simplicial assumption, a, b, and d do not havedifferent signs, so we can choose them to be non-negative.
8 A Needlessly Long Proof Reducing the 1−0− 0− 0 Case to the 1-0-0-0 and c = 1 Case
Let M c =
(1 0 0 c0 a b d
).
Definition 8.1. Let MIN c denote the set of minimal elements in eachresidue class for M c
N.
Lemma 8.2. Let M c =
(1 0 0 c0 a b d
)and M c+i =
(1 0 0 c + i0 a b d
)with b2 =
(cd
)and b′2 =
(c+id
). If m = c1b1 + c2b2 and m′ = c1b1 + c2b2, then
m ∈MIN c ⇐⇒ m′ ∈MIN c+1.
Proof:[⇐] It is equivalent to show that if m 6∈MIN c, then m′ 6∈MIN c+i.
m 6∈ MIN c ⇒ ∃ w ∈ MIN c, with w < m and w ≡ m mod A.Let w = e1b1 + e2b2. Consider w′ = e1b1 + e2b2 + e2
(i0
)= e1b1 + e2b
′2.
w′ ≡ w ≡ m ≡ m′ because the addition of a1 does not change the residueclass and w ≡ m.
25
e2 ≤ c2 because (w)x ≤ (m)x because w < m and b1 has no x component.e2 ≤ c2 ⇒ e2
(i0
)≤ c2
(i0
). Hence,
w′ = e1b1 + e2b2 + e2
(i0
)<
< c1b1 + c2b2 + e2
(i0
)≤
≤ c1b1 + c2b2 + c2
(i0
)=
= c1b1 + c2b′2 = m′.
w′ ∈M ′N because w′ = e1b1 + e2b
′2, so w′ < m′ ⇒ m′ 6∈MIN c+i.
[⇒] It is equivalent to show that if m′ 6∈MIN c+i, then m 6∈MIN c.
m′ 6∈ MIN c+i ⇒ ∃ w′ ∈ MIN c+i, with w′ < m′ and w′ ≡ m′ mod A.Let w′ = e1b1 + e2b
′2 = e1b1 + e2b2 + e2
(i0
). Consider w = e1b1 + e2b2.
w′ ≡ w ≡ m ≡ m′ because the addition of a1 does not change the residueclass and w′ ≡ m′.
Claim: e2 ≤ c2. (w′)x ≤ (m′)x, and because the x coordinate is inde-pendent of b1 and only dependent on b′2, if (w′)x ≤ (m′)x, then e2 ≤ c2. Soe2b2 ≤ c2b2, so (w)x ≤ (m)x.
Now (m)y = (m′)y ≥ (w′)y = (w)y, so (m)y ≥ (w)y.
Now for m′ and w′ if (m′)y = (w′)y, then (m′)x > (w′)x, and if (m′)x =(w′)x, then (m′)y > (w′)y. If not, m′ = w′, so m′ ∈MIN c+i.
So we have that(m)y > (w)y and (m)x = (w)x,(m)y = (w)y and (m)x > (w)x or(m)y > (w)y and (m)x > (w)x.
In any case, w < m, w ≡ m mod A and w ∈MN. So m 6∈MIN c.
Lemma 8.3. Consider M c and M c+i as above. Then for any gi ∈ G(M c),there is some g′i ∈ G(M c+i) such that if gi + A1 = lubwk, then ∃ g′i withgi + A1 = lubw′
k and vice versa. Hence, |G(M c)| = |G(M c+i)|.
Proof: Consider gi ∈ G(M c). gi + A1 = lubwk, with wk ⊂ MIN c.For each wk, ∃w′
k ∈ MIN c+i with wk ≡ w′k by lemma 8.2. Consider w′
k;
26
w′k is a complete set of coset representatives and w′
k ⊂ MIN c+i, soh = lubw′
k is g complete.
Suppose towards contradiction that h is not a g vector. Then ∃ g?, withg? < h, and g? complete. Then g? + A1 = lubu′k for some u′k ⊂MIN c+i.
Consider ∃ uk ⊂MIN c. uk is a complete set of coset representatives,and lubuk < lubwk because lubu′k < lubw′
k ⇒ gi 6∈ G(M), which isa contradiction. So h = lubw′
k − A1 ∈ G(M c+i).We can follow a similar argument to show that if there is some g′j ∈ G(M ′)
there must be a corresponding gj ∈ G(M c). Hence, |G(M c)| = |G(M c+i)|.
Theorem 8.4. Let the submonoid M0 = [Ab1] generate some frobenius vec-tor, g0. Let c=1. Then (gi)x = i− 1 for any gi ∈ G(M c).
Proof: Because b1 is along the y axis, we know that G(M0) = |A|b1 −A1 = |A|
(0b
)−(1a
)=( −1|A|b−a
)=( −1
a(b−1)
). So g0 =
( −1a(b−1)
).
Consider b2 =(1d
). If b2 6∈MIN then by lemma 6.3 g0 = G(M c).
So b2 ∈ MIN ⇒ b2 ≡ jb1 with b2 incomparable to jb1. Likewise,hb1 + b2 ∈MIN for 0 ≤ h ≤ |A| − j = a− j.
There are |A| = a equivalence classes, with the jth through ath repre-sented by hb1 + b2 and the first through j − 1th represented by kb1. Let theunion of the two sets, hb1 + b2|0 ≤ h ≤ a − j ∪ kb1|1 ≤ k ≤ j − 1 bedenoted wk. Then lubwk − A1 = γ is g-complete.
Likewise, there is no g? ∈ G(M c) with g? < γ because then (g?)x = 0or −1. But the only g vector with (g)x = −1 is not less than γ and γ isminimal for an x coordinate of 0 by construction. So γ is a g vector, g1 with(g1)x = 1− 1 = 0.
Suppose that you have two consecutive g vectors, gi and gi+1 with (gi)x =i − 1. Suppose towards contradiction that (gi)x < (gi+1)x and (gi)x + 1 6=(gi+1)x.
27
gi + A1 = lubuk with some uk with uk = αb1 + (i − 1)b2 because(gi)x = (i − 1). However, by lemma 6.4, uk ∈ MIN ⇒ (i − 1)b2 ∈ MIN .gi+1 ∈ G(M c)⇒ sb2 ∈MIN for some s > i− 1.
However, for any value of s, by lemma 6.4, ib2 ∈ MIN . Then if ib2 ∈MIN , ib2 ≡ j′. By a similar argument to the one above, h′b1+ib2 ∈MIN for0 ≤ h′ ≤ |A|−j′. Then we have some w′
k say with w′k = c1b1+c2b2|w′
k ∈MIN, c2 ≤ i. By a similar argument to γ ∈ G(M c), lubw′
k − A1 = g? ∈G(M c). But (gstar)x < (gi+1)x ⇒ gi and gi+1 are not consecutive. This is acontradiction. Therefor, (gi)x + 1 = (gi+1)x.
So by the principle of mathematical induction, our claim holds.
Lemma 8.5. If there is some gi ∈ G(M1), then there is g′i ∈ G(M c) suchthat g′i = gi + i
(c−10
).
Proof: Let gi + A1 = lubw for some w ⊂ MIN1. Let we be theset of elements of w with the greatest x coordinate. We then have that(lubw)x = (we)x = i by lemma 8.4.
We have from lemma 8.2 that for each w ∈ MIN1, there is some w′ ∈MIN c such that if w = c1b1 +c2b2, then w′ = c1b1 +c2b2 +c2
(c−10
). By lemma
8.3 there is some g′i ∈ G(M c) with g′i + A1 = lubw′.
(lubw)y = (lubw′)y because c2
(c−10
)does not change the y coordinate.
If we has a maximum x coordinate for w, then w′e will for w′. w′
e =we + c2
(c−10
)= we + i
(c−10
)because c2 = i by lemma 8.4. So (lubw′)y =
(lubw)y and (lubw′)x = (lubw)x + i(c− 1). Therefor g′i = gi + i(
c−10
).
.
Theorem 8.6. Suppose M1 =
(1 0 0 10 a b d
)and M c =
(1 0 0 c0 a b d
)and G(M1) = g0, g1, . . . , gk. Then G(M c) = gj + j
(c−10
) for 0 ≤ j ≤ k.
Proof: By lemma 8.3, for any gj ∈ G(M1) there is some g′j ∈ G(M c)
and vice versa. By lemma 8.5 g′j = gj + j(
c−10
).
28
9 All bi Lying on Boundary
Define a “prime” vector as an integer vector whose coordinates have gcd 1. Itis evident that we may express any integral vector as a product of an integerscalar and a prime vector
Theorem 9.1. Let all the prime vectors that define the A-cone as α1, α2, . . . , αn
(ie, all A-vectors can be expressed as zi,jαi, where zi,j is an integer scalarand i, j are integers). If all the generating vectors can be expressed as integerscalar multiples of the αi, then if there is a frobenius vector, it is unique andequal to F =
∑ni=1 fjαj, where fi is the frobenius number of all the zi,j, where
j varies. If, for any αi fi does not exist, then no frobenius vector exists.
Proof. If one of the fi does not exist, wlog f1, this means that all the gener-ating vectors with prime vector α1 may be expressed as mjkα1, for positiveintegers mj, and some positive integer k 6= 1. It is evident that any integralvector whose α1-component is an integer not congruent to 0 mod k is notin the monoid- hence no frobenius vector can exist.
Assume there exists a complete vector, F ′ which is incomparable orsmaller than F . Express F ′ as [c1c2...cn] · [αiα2 . . . αn] (the ci are positive,but not necessarily integers). For at least one i, ci < fi. Wlog, let c1 < f1.Consider the vector [(f1)(c2 + 1)(c3 + 1)...(cn + 1)] · [αiα2 . . . αn]. This vectoris not in the monoid- the term f1α1 cannot be expressed as a sum of thegenerating vectors who have α1 as their prime vector, by the definition of f1.Thus we have a vector larger than F ′ that is not in the monoid. We claimthat the gcd of the generating vectors is equivalent to det[α1α2 . . . αn]. Saythat pd is the highest power of some prime p to divide the gcd. Say thatpe is the highest power of p to divide det[α1α2 . . . αn]. Select vectors wherenone is a scalar multiple of the other; that is, one multiple each of each αi.The determinant of any other collection of vectors will have one vector arational scalar multiple of the other, and that determinant is 0 and may bedisregarded in the gcd. It is possible to make a selection such that none ofthe scalar multiples is a multiple of p. Otherwise, if this is impossible for acertain αi, wlog say α1, then all generating vectors that are a multiple of α1
also have a scalar factor of p, so f1 does not exist. Hence d = e.If the gcd of the generating vectors is 1 then the determinant of the αi is 1-
meaning that every integral lattice point can also be expressed as an integrallattice point of the αi-basis. But by the definition of the frobenius number,every vector whose αi component exceeds fi for every i is representable in the
29
monoid. Thus every vector in the αi lattice larger than F is in the monoid.Since every integral lattice point larger than F corresponds to vector in theαi lattice larger than F , it follows that F is a complete vector. Since we alsoknow that every vector smaller than or incomparable to F is not complete,it follows that F is the unique Frobenius vector.
10 Selmer Lattice: Properties and Applica-
tions
The Selmer Lattice is a construct invented by Ernst Selmer to solve the 1-dimensional Frobenius problem in 3 generators. In this paper we generalizethe tools he uses to the vector case, in an attempt to learn more about thehigher-dimensional analogues of the 1-dimensional problem.
10.1 Properties
10.1.1 Definitions
1. We are considering the set of points xb1 + yb2 where x and y are non-negative integers. We arrange the elements in a cartesian lattice in thefirst quadrant, where the x-axis represents b2 and the y-axis representsb1.
2. We have the relation x1b1 + x2b2≥By1b1 + y2b2 if xi ≥ yi for i = 1, 2.We refer to this relation as being “B-greater”.
3. The zero vector is the trivial zero.
4. A zero is a vector that is congruent to the zero vector.
5. An element is insignificant if it is≥B than a nontrivial zero. An elementthat is not insignificant is significant.
6. We define the Selmer Lattice to be the set of significant vectors.
7. The element 1b1 + 0b2 belongs to the residue class 1. The residue classs is the residue class where 0b1 + 1b2 belongs.
30
8. Given ib1 + jb2, we define it’s contained set to be the set of elementsin the Selmer diagram that are ≤Bib1 + jb2.
9. We call an element of the Selmer diagram good when it’s contained setcontains a complete residue set mod |A|.
10. Define the Minimal Selmer Diagram to be the diagram including onlythe elements in MIN .
11. v ≡> w means that the vectors v and w are congruent and that v isgreater or equal than w.
12. v≡<w means that the vectors v and w are congruent and that v is lesseror equal than w.
13. v‖≡w means that the vectors v and w are congruent and that v is in-comparable to w.
14. Let g ∈ G. We know that there exist a complete set of residues N ∈MIN such that lub(N) − A1 = g. Now add the constraint that N iscontained in the first k rows of the Selmer lattice. Let m be the minimalvalue of k for which such an N exist. We now define Low(g) = m andsay N is a minimal representation for g if it is contained in the firstLow(g) rows of the Selmer lattice.
Proof. Since x1b1+y1b2≥Bx2b1+y2b2, we have x1b1+y1b2-x1b1+y1b2=(n1, n2)for some nonnegative integers n1, n2. Thus x1b1+y1b2-x1b1+y1b2=n1b1+n2b2.Given that b1 and b2 can both be expressed as positive sums of the a-vectors,we know that x1b1 + y1b2-x1b1 + y1b2 can be expressed as positive sums ofthe a-vectors. Thus x1b1 + y1b2 ≥ x2b1, y2b2
31
Theorem 10.2. If (x1, y1)≥B(x2, y2) for some (x2, y2) in the first quadrantof the Selmer Lattice, then (x1, y1) is not significant
Proof. (x1, y1)≥B(x1, y1)− (x2, y2), which is a zero.
Theorem 10.3. All points in the MIN set are significant.
Proof. Given xb1 + yb2 an insignificant point. There exists a nontrivial zeroz1b1 + z2b2≤Bxb1 + yb2. Thus, (x− z1)b1 + (y − z2)b2 is a point on the firstquadrant of the Selmer Lattice. Since z1b1 +z2b2 ≡ 0, (x−z1)b1, (y−z2)b2 ≡xb1 + yb2. Also, since (x− z1)b1 + (y − z2)b2 + z1b1 + z2b2=xb1 + yb2, xb1 +yb2≥B(x−z1, y−z2) and by Theorem 10.1 (x−z1)b1 +(y−z2)b2 ≤ xb1 +yb2.In fact (x − z1)b1 + (y − z2)b2 < xb1 + yb2 since z1b1 + z2b2 is a nontrivialzero. Thus (x− z1)b1 + (y − z2)b2 ≡< xb1 + yb2 and so xb1 + yb2 is not in theMIN set.
Theorem 10.4. Order the nontrivial significant zeroes according to increas-ing b2-coordinate. The zeroes will then also be ordered according to decreasingb1-coordinate.
Proof. We prove this by induction. (|A|, 0) is the first element in this or-dering. Consider the Zk, the kth zero of the ordering. It has a larger b2
coordinate than the Zk−1 zero. If Zk’s b1-coordinate were greater than orequal to Zk−1’s b1 coordinate, Zk would be B-greater than Zk−1, contradict-ing its significance.
Note that this proves that each row of the Selmer Lattice of significantelements contains no more elements than the row above it.
10.1.3 Wavy Cuts
Theorem 10.5. Given two significant zeroes, z1b1 + o1b2 and z1b1 + o1b2 iff(z1, o1) ≤ (z2, o2) and o1 < o2, all elements of the MIN set have b2-coordinateless than o2 − o1. Iff (z1, o1) ≤ (z2, o2) and o1 > o2 then all elements of MINhave b1-coordinate less than z2 − z1.
32
Proof. Say o1 < o2. Given any xb1 + yb2 with y ≥ o2 − o1, we know thatxb1 + yb2 ≡ (z1b1 + o1b2)− (z2b1 + o2b2) + (xb1 + yb2). This is true becausez1b1 + o1b2 and z1b1 + o1b2 both belong to residue class 0. (z1b1 + o1b2) −(z2b1+o2b2)+(xb1+yb2) is a point on the first quadrant of the Selmer Latticebecause y ≥ o2 − o1, and since o2 > o1, z2 < z1 (Theorem 10.4). But since(z2b1 + o2b2)− (z1b1 + o1b2) > 0, we have (z1b1 + o1b2)− (z2b1 + o2b2)+ (xb1 +yb2) < (xb1 + yb2) and so (z1b1 + o1b2)− (z2b1 + o2b2)+ (xb1, yb2) ≡< xb1 + yb2.Thus xb1 + yb2 is not in MIN.
If instead o1 > o2, then z1 < z2 (Theorem 10.4). Given any xb1 + yb2
with x ≥ z2− z1, we know that (x, y) ≡ z1b1 + o1b2− z2b1 + o2b2 + xb1 + yb2.This is true because z1b1 + o1b2 and z1b1 + o1b2 both belong to residue class0. (z1b1 + o1b2) − (z2b1 + o2b2 + xb1 + yb2 is a point on the Selmer Latticebecause x ≥ z2 − z1 and o1 > o2. But since z2b1 + o2b2 − (z1b1 + o1b2 > 0,we have (z1b1 + o1b2) − (z2b1 + o2b2) + (xb1 + yb2) < (xb1 + yb2) and so(z1b1 + o1b2)− (z2b1 + o2b2) + (xb1 + yb2) ≡< xb1 + yb2. Thus xb1 + yb2 is notin MIN.
Theorem 10.6. Given two congruent significant elements (x1, y1) and (x2, y2)where x2 > x1 there exists a zero (z, o) where z, o ∈ [0, |A|] and where(x2, y2)− (x1, y1) = (|A|, 0)− (z, o)
Proof. Since x2 > x1 we have also y2 < y1, otherwise (x2, y2) cannot besignificant, by Theorem 10.2. (x1, y1) − (x2, y2) ≡ 0 we have also (|A|, 0) +(x1, y1) − (x2, y2) ≡ 0. Also, since x2 > x1 and y2 < y1 (|A|, 0) + (x1, y1) −(x2, y2) ≡ 0 lies in the first quadrant and has b1-value less than |A|. Thus,we need only set (z, o) = (|A|, 0) + (x1, y1)− (x2, y2)
Theorem 10.7. Let (z1, o1) be a zero other than (|A|, 0) such that (z1, o1) ≡<(|A|, 0) and o1 is minimal. Let (z2, o2) be a zero other than (|A|, 0) such that(z2, o2) ≡> (|A|, 0) and |A| − z2 is minimal. The MIN set is comprised of allsignificant elements with b1-value less than |A|−z2 and b2-value less than o1.
Proof. By Theorem 10.5 we know that all elements with b1-value greater thanor equal to |A| − z2 and b2-value greater than or equal to o1 cannot be inthe MIN set. It remains to show that the significant elements that are left
33
comprise the MIN set. It suffices to show that any two congruent significantelements in this reduced set are incomparable. Suppose instead there exist apair of significant elements, (x1, y1), (x2, y2) in this set that are comparable.Wlog let x2 > x1. Then also y2 < y1, otherwise (x2, y2) is insignificant byTheorem 10.2. We have both x2−x1 < |A|−z2 and y1−y2 < o1. By Theorem10.7 there exists (z, o) such that (|A|, 0) − (z, o) = (x2, y2) − (x1, y1).Since(x2, y2), (x1, y1) are comparable, it must follow that (x2, y2)−(x1, y1) equals asum of A-vectors with either all positive or all negative coefficients. (|A|, 0)−(z, o) must equal that same sum, and so (|A|, 0), (z, o) are comparable. But|A|−z = x2−x1 < |A|−z2 and o = y2−y1 < o1, contradicting the minimalityof either |A| − z2 or o1.Hence, given any congruence class, we know all its minimal elements mustlie in this minimal Selmer Lattice, and we know that all elements of thiscongruence class in this reduced set are mutually incomparable. Thus theyare all minimal.
We want to find the (z1, z2) ≡> (|A|, 0) where z2 is minimum.
Theorem 10.8. In the Selmer Lattice bounded by b1 and b2-values ∈ [0, |A|],there exists one zero a row because b1 is a generator. The zeroes are (|A|−(ismod |A|), i) where i ranges from 0 to |A|.
Proof. Since sb1 ≡ b2, we know that for any zero Z, Z+(−s, 1) is also a zero.Also, for any zero Z, Z + (|A|, 0) is a zero. Since |A|, 0) is a zero, for theb2 = i row the element (|A| − is, i) is a zero. If |A| − is is negative we canadd (|A|, 0) to (|A| − is, i) enough times so the b1-coordinate lies in [0, |A|).This will imply (|A| − (is mod |A|), i) is a zero, as we required.
Theorem 10.9. There exists two real numbers, µ and ν such that for anytwo positive integers c and d so
1. cd≤ ν ↔ (x, y) < (x, y) + (−c, d),
2. µ > cd
> ν ↔ (x, y) is incomparable to (x, y) + (−c, d)
3. cd≥ µ↔ (x, y) > (x, y) + (−c, d)
for all elements (x, y).
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Proof. It suffices to prove this assertion true for (0, 0). This is because (0, 0)−(−c, d) = (x, y)− ((x, y)− (−c, d)) for all x, y and thus the relation between(0, 0) and (−c, d) would be the same as the relation between (x, y) and (x, y)−(−c, d).Define µ as the smallest real number so b2 < µ · b1. Define ν as the largestreal number so b2 > ν · b1. If c
d≤ ν then
−cb1 + db2 = d(− c
db1 + b2) ≥ d(−νb1 + b2) > c(−b2 + b2) = 0
If (−c, d) > (0, 0) then −cb1 + db2 > 0 and b2 > cdb1. Since ν is the largest
number that b2 > νb1, we must have cd≤ ν. Thus c
d≤ ν ↔ (−c, d) > 0.
If cd≥ µ then
−cb1 + db2 = d(− c
db1 + b2) ≤ d(−µb1 + b2) < c(−b2 + b2) = 0
If (−c, d) < (0, 0) then −cb1 + db2 < 0 and b2 < cdb1. Since µ is the smallest
number that b2 < µb1, we must have cd≥ µ. Thus c
d≥ µ↔ (−c, d) < 0.
(−c, d) = 0 is impossible, otherwise db2 = cb1 and so b2 is a scalar multipleof b1, a case we are not considering. Hence if (−c, d) ≮ 0 and (−c, d) ≯ 0then (−c, d) must be incomparable to 0. By elimination we have µ > c
d>
ν ↔ (x, y) is incomparable to (x, y) + (−c, d)
If we want to find the “height” of the MIN set then we must identify thezero in (|A| − (is mod |A|), i) so that (is mod |A|)
i≤ ν and i is minimal.
The zero that i refers to will be the zero larger than (|A|, 0) with the minimalb2-coordinate.
We use a continued fraction algorithm to determine the proper i. Set s−1
as |A| and s0 as s and we have
s−1 = q1s0 − s1 0 ≤ s1 < s0
s0 = q2s1 − s2 0 ≤ s2 < s1
s1 = q3s2 − s3 0 ≤ s3 < s2
· · ·sk−2 = qksk−1 0 ≤ sk < sk−1 (1)
sk−1 = qk+1sk 0 = sk+1 < sk
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If s0 = 0 we set m = −1. We also define integers Pi by P−1 = 0, P0 = 1and Pi+1 = qi+1Pi − Pi−1, i = 0, 1, 2, . . . k.
Since qi ≥ 2 it follows by induction that Pi + 1 > Pi. Also si > si + 1.Hence
0 =sk+1
Pk+1
<sk
Pk
< . . . <s0
P0
<s−1
P−1
=∞ (2)
There are unique integers λ and ξ such that sλ+1
Pλ+1< µ ≤ sλ
Pλand
sξ
Pξ≤ ν <
sξ−1
Pξ−1.
Theorem 10.10. Pi+1s0 ≡ si+1 mod |A| for all i ∈ [0...|A| − 1]
Proof. We induct on i. We have P1 = q1P0 − P−1 = q1. Thus P1s0 = q1s0 =|A|+ s1 ≡ s1 mod |A|. Also, similarly
Assume this proposition holds up to i. Pi+1s0 = (qi+1Pi−Pi−1)s0 ≡ qi+1si−si−1 = si+1 by the induction hypotheses and (1).
Theorem 10.11. Among the integers s0, 2s0, . . . , (Pi − 1)s0 taken modulo|A|, the smallest among these is Pi−1s0 mod |A|.
Proof. Using the fact that Pi−1s0 = si < sj = Pjs0 for i < j, by Theorem10.10,we prove this proposition by induction. This assertion is trivially truefor i = 1. P0 = 1, and so s0, 2s0, . . . (P1 − 1)s0 are all less than |A| =q1s0 − s1 = P1s0 − s1. Thus s0 is the smallest among these mod |A|. Letus assume the assertion is true for some i = k − 1. For the equation js0
mod |A|, if j < Pk − 1 then using the induction hypothesis
Pk−1s0 mod |A| < Pk−2s0 mod |A| ≤ js0 mod |A| (3)
If j > Pk−1 then we may express j as qPk−1 − r for positive integers 2 ≤ qand 0 ≤ r < Pk−1. We claim that j = Pk, so js0 = Pks0 = (qkPk−1−Pk−2)s0
is the smallest value of js0 so js0 < Pk−1s0, taken mod |A|. Assume thatthere is a smaller j = j′ that fulfills js0 < Pk−1s0 taken mod |A|. Sayfirst that j′ = qkPk−1 − r, where r > Pk−2. If qkPk−1s0 − rs0 < Pk−1s0
mod |A|, because we have qkPk−1s0−Pk−2s0 < Pk−1s0 mod |A| as well then
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we must have rs0 − Pk−2s0 < Pk−1s0 in mod |A|; recall that rs0 > Pk−2s0
in mod |A| by the induction hypothesis. But rs0 − Pk−2s0 = (r − Pk−2)s0,and (r − Pk−2) is less than Pk−1. But we know that if n is any integer lessthan Pk−1, Pk−1s0 < ns0 mod |A| using (3).Consider instead j′ = qPk−1−r where q = qk−t, where t is a positive integer.If r = Pk−2, we can see that j′s0 ≡ Pks0 − tPk−2s0. Note that tPk−1s0 mustlie between Pk−1s0 and |A|. This is because
Thus, j′s0 ≡ Pks0− tPk−1s0 mod |A| > Pks0 mod |A| (recall that Pk−1s0 >Pks0 mod |A|). If r 6= Pk−2 then necessarily Pk−2s0 < rs0 < |A|. Thusj′s0 = qPk−1s0 − rs0 lies between qPk−1s0 − Pk−2s0 mod |A|, and qPk−1s0
mod |A| (both greater than Pks0 mod |A|). Thus j′s0 mod |A| cannot beless than Pks0 mod |A|. Hence the smallest possible j where js0 is less thanPk−1 is Pk
Theorem 10.12. The MIN set is the set of significant elements with b1-valueless than sλ and b2-value less than Pξ.
Proof. We use Theorem 10.7 and look for the zero (z, w) where (|A|, 0) ≡<(z, w) and w is minimum. By Theorem 10.9 we must have |A|−z
w≤ ν so we
are looking for the zero with the smallest w that fulfills this condition. Weclaim that this zero has w = Pξ.
By Theorem 10.8 |A|−zw≤ ν is equivalent to (ws mod |A|)
w≤ ν. We know that
Pws0 ≡ sw by Theorem 10.12. Thus (ws mod |A|)w
≤ ν is equivalent to sw
Pw≤ ν
(remembering that s0 = s) , which is true for w = Pξ.Furthermore, no w less than Pξ will fulfill this condition. It is evident that noPi with i < ξ fulfills this condition, otherwise we have si
Pi≤ ν, contradicting
ν <sξ−1
Pξ−1. Also, no w between some Pi+1 and Pi can fulfill this condition.
For this, it suffices to show that Pis0 mod |A| is the minimum value of js0
mod |A| for j = 1, 2 . . . Pi+1 − 1. This is proven in Theorem 10.11. Henceif a value w lies between some Pi − 1 and Pi with i < ξ we have ws > Pismod |A| and so (ws mod |A|)
w> (Pis mod |A|)
Pi> ν. Thus Pξ is the smallest value
of w where the zero on the row b2 = w is larger than (|A|, 0). An analogousargument will show that sλ will be the smallest element z where zero incolumn |A| − z is smaller than (|A|, 0).
37
Theorem 10.13. The important zeroes are (si−1 − ksi, kPi − Pi−1) for iranging from -1 to and k ranging from 1 to qi.
Proof. We know that the first q1 rows have significant zeroes.(|A|, 0) is obvi-ously an important zero, and since this list of the first q1 zeroes has decreasingb1-value and increasing b2-value, all of them are important. Say, for some ithat the list of all important zeroes before the row Pi−Pi−1 is given as in theproposition. We wish to show that the sequence of important zeroes betweenrow Pi − Pi−1 and row Pi+1 − Pi is (si−1 − ksi, kPi − Pi−1), where k rangesfrom 1 to
10.1.4 Other Important Properties
Theorem 10.14. If row i has fewer elements than row i−1 for 1 ≤ i ≤ |A|,then the last element of row i is |A| − 1.
Proof. Given a row i that has fewer elements than row i−1, let row i containp elements and let row i−1 contain p+q, where p, q ∈ N. Since ib1+(p+1)b2
is not in the Selmer diagram, there must be an element within the first p+1rows and i columns of the Selmer diagram that has the same congruenceclass as it. Also, since ib1 + pb2 is in the Selmer diagram, there is no otherelement within the first p rows and i columns that is congruent to it.
This means that the element congruent to the ib1 + (p + 1)b2 withinthe first p + 1 rows and i columns of the Selmer diagram must be in thefirst column - otherwise, the element in the same row and one column be-fore it is congruent to ib1 + pb2. Furthermore, the element congruent toib1 + (p + 1)b2 must be in the first row, otherwise (i − 1)b1 + (p + 1)b2 iscongruent to something else within the first i − 1 rows and p + 1 columns,which contradicts our assumption. Thus, ib1 + (p + 1)b2 ≡ 0 mod |A|, andso ib1 + pb2 ≡ p− 1 mod |A|.
Analogously, if column i contains fewer elements than column i− 1, thenthe last element of column i ≡ p− s mod |A|.
Theorem 10.15. If there is a Frobenius vector g such that g + A1 ≥ somegood element v, then g + A1 = v.
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Proof. Recall that ∃ω1, ..., ω|A| ∈ MIN , a complete set of residue classes,such that g + A1 = lub(ω1, ..., ω|A|). But since v is ≥ a complete set ofresidue classes, we can choose these elements to take the lub of, which willyield v. Thus v−A1 is a complete vector, so g+A1 cannot be greater thanv.
As a consequence of this theorem, for any good vector v, if v − A1
is not a Frobenius vector, then 6 ∃ vectors ω1, ..., ω|A|−1 ∈ MIN such thatlub(ω1, ..., ω|A|−1,v)−A1 is a Frobenius vector. It is also easy to see that ifsome good element is in
Theorem 10.16. if row i contains more elements than row i + 1 (for 1 ≤i ≤ |A| − 1), then the last element of row i is good.
Proof. Assume that row i + 1 contains p elements, and that row i containsp + q elements, where p, q ∈ N. To show that (i, p + q) is good, it is easy tosee that it suffices to show that every number of the form (i+1)b1 + kb2 for1 ≤ k ≤ p + q is congruent to some element in the contained set of (i, p + q).
By property 1, we know that (i + 1)b1 + (p + 1)b2 ≡ 0 mod |A|, and soall elements of the form (i + 1)b1 + (p + j)b2 ≡ j − 1 for 1 ≤ j ≤ q. It isclear that each of these elements is congruent to something in the containedset of (i, p + q) (in fact, they correspond to the first q elements of the firstrow).
Also by property 1, we have that the element (i + 1, p) ≡ |A| − 1, fromwhich it follows that the elements (i + 1, 1), (i + 1, 2), ...(i + 1, p) ≡ |A| − p +1, |A| − p + 2, ...|A| − 1. But by property 1, we know that the first row thatcontains exactly p + q elements ends with |A|, and so this row must containelements of the congruence classes (i + 1, 1), (i + 1, 2), ...(i + 1, p). Since thisrow is clearly in the contained set of (i, p + q), we obtain the desired result.
Analogously, if column i contains more elements than row i + 1 (for 1 ≤i ≤ |A| − 1), then the last element of column i is good.
Theorem 10.17. For the case where there are two B-vectors b1, b2, we labelthem so b1 is not greater than b2. If gcd(A|b1) is 1 (as we are assuming),there exists an integer 0 ≤ s < |A| so s · b1 ≡ b2 mod A
39
Proof. The gcd of the matrix (A|b1) is 1. Thus for every congruence classof A, there will be an integer n which is at most |A| where n · b1 is in thatcongruence class. Thus there must exist s < |A| so s · b1 ≡ b2 mod A
10.2 Applications
10.2.1 Solving the m = 2 Case
In this section we attempt to generalize the partial solutions to the 3 gen-erator Frobenius problem found in the papers “On the Linear DiophantineProblem of Frobenius” (1977) and “‘On the Linear Diophantine Problem ofFrobenius in Three Variables”(1978). we are looking at the cases governedby the conditions (q +1)a2 ≥ (s− r)a1 and (qm+1)a2 ≥ pa1 in the notationof those two papers. Generalizing the first case to more than one dimensionis fairly straightforward; generalizing the second is a bit more complicated.In all the following we will be assuming sb1 ≡> b2; the case where sb1‖≡ b2 hasbeen completely solved, and we can choose to number b1, b2 so sb1 ≡< b2 isimpossible. By an earlier result, we can, and do assume that both b1 and b2
are generators.
Theorem 10.18. For the case where there are two B-vectors b1, b2, we labelthem so b1 is not greater than b2. If gcd(A|b1) is 1 (as we are assuming),there exists an integer 0 ≤ s < |A| so s · b1 ≡ b2 mod A
The gcd of the matrix (A|b1) is 1. Thus for every congruence class ofA, there will be an integer n which is at most |A| where n · b1 is in thatcongruence class. Thus there must exist s < |A| so s · b1 ≡ b2 mod Asb1 ≥ b2
Theorem 10.19. Given four vectors where the gcd of (A|b1) is 1 and theinteger s is defined as in Theorem 10.17. We define q,r such that |A| = qs+r,where 0 < r ≤ s. If s·b1 ≥ b2 and (q+1)b2 ≥ (s−r)b1 then |MIN | = |A| andthere exists only one Frobenius vector, lub((r−1)b1+qb2, (s−1)b1+(q−1)b2)-A.
40
We use the Selmer Lattice. As shorthand for x·b1+y·b2, we write (x,y).We mark the MIN set on the Selmer Lattice. Given the conditions set outin the statement of the theorem, we claim that all the elements of the MINset are in the following table:0 b1 2b1 . . . (s− 1)b1
Consider any point (x, y). This point represents xb1 + yb2. If x ≥ s thenxb1 +yb2 ≡> (x−s)b1 +(y+1)b2. Thus any point with an x-value greater thans-1 cannot be minimal in its residue class. Also, there are no elements of MINobtained by extending the last line to the right. Since the diagram containsrepresentatives of qs + r = |A| obtained from each other by addition of b1,rb1 + qb2 ≡ (qs+ r)b1 ≡ 0 mod A, so we have rb1 + qb2 ≡> 0. Thus extendingthe last line to the right will get us s-r elements congruent to and larger thanthe first s-r elements in the first row. Thus, (q+1)b2 ≡ sb1 +(qb2 ≡ (s−r)b1.This means there will also be no elements of MIN with y-coordinate largerthan q. By (q + 1)b2 ≡> (s − r)b1, we know that any point (x, y) with y > qwill not be minimal since xb1 + yb2 ≡> (x + s − r)b1 + (y − q − 1)b2 . Weknow that every residue class is represented in this diagram since the points,read from left to right and then from top to bottom ,are congruent to 0,b1, 2b1, . . . (|A|− 1)b1, due to the fact that s · b1 ≡ b2 mod A. Thus these theentire MIN set is expressed by this diagram. All the points in the diagram aresmaller than (r-1,q) or (s-1,q-1). This is because we may obtain one of thesetwo points from any point in the diagram by adding some b1 and b2 vectors.The Frobenius vector is thus lub((r − 1)b1 + qb2, (s− 1)b1 + (q − 1)b2)-A.
Corollary : When s·b1 ≥ b2 and s divides |A|, the unique Frobenius vectoris (s− 1)b1 + (q − 1)b2 − A.
If s s divides |A| then we just set r = s in Theorem 10.19. This yields usthe answer lub((s−1)b1 +qb2, (s−1)b1 +qb2)-A. Since (s−1)b1 +(q−1)b2)-A≥ (s− 1)b1 + (q − 1)b2)-A the frobenius vector is (s− 1)b1 + (q − 1)b2-A.
Theorem 10.20. Given four pairwise relatively prime vectors and the integers defined in Theorem 10.17. We define q,r,m,p such that |A| = qs + r ands = mr + p. If s · b1 ≥ b2 and (qm + 1)b2 ≤ pb1, the set lub((s − 1 −(i− 1)r)b1 + (q − 1)b2, (r − 1)b1 + iqb2)− A where i varies from 1 to m+1
41
only contains complete vectors. Furthermore, all of the Frobenius vectors arecontained within this set.
Define S to be the set of points where x ∈ [0, r − 1]and y ∈ [0, q − 1].We define for natural number i the sets Ai as S + (p + mr− ir, 0) and Bi asS + (0, 1 + qi). Furthermore, define
αi = (s− 1− (i− 1)r)b1 + (q − 1)b2
βi = (r − 1)b1 + (i + 1)qb2
Note that αi ∈ Ai and βi ∈ Bi. The congruence αi ≡ βi is a conse-quence of s · b1 ≡ b2 mod A and |A| = qs + r. This is because αi − βi =b1(s− ir)−b2(iq+1) ≡ −b1ir−b2iq ≡ −i(qs+r)b1 ≡ 0 Also, we note that αi
and βi are the members of Ai and Bi respectively where the x and y valuesare maximum. Thus, any point in Ai may be expressed as αi − (X, Y ) forsome X, Y where X is in [0, r − 1] and Y is in [0, q − 1]. βi − (X, Y ) is apoint in Bi that is congruent to it. We can see that corresponding elementsof Ai and Bi are congruent.
(r, q) is congruent to 0 since rb1 + qb2 ≡ (qs + r)b1. We have s = rm + p,p = s − rm and so pb1 ≡ (s − rm)b1 ≡ b2 − rmb1 ≡ (qm + 1)b2 (since−rb1 ≡ qb2). Thus (qm + 1)b2 ≡> pb1
Any point (x, y) in the lattice with x ≥ q and y ≥ r cannot be min-imal by Theorem 10.3. Since we have the condition (qm + 1)b2 ≡> pb1,(r, q) ≡< (r + qm + 1, q − p) and by Theorem 10.5 any point (x, y) withy ≥ qm+1 cannot be in MIN. Thus we need only consider Bi up to i = m−1.
Furthermore, given Ai, Aj where 1 ≤ i < j ≤ m no element in Ai is congruentto an element in Aj. This is because all the elements of Ai and Aj lie withinthe bounds y < q, x < s which is a subset of the set of points referenced inTheorem 10.19. The set of points in Theorem 10.19 all belonged to differentcongruence classes, thus the set of points in Ai and Aj all belong in differentcongruence classes as well. Also, within any Ai or Aj distinct points belongto distinct congruence classes. Hence the points in the sets A1, A2 . . . Am allbelong to distinct congruence classes. Since the points in Ai are congruent
42
to the points in Bi, it is also true that the points in B1, B2 . . . Bm all belongto distinct congruence classes
We know all qs+r congruence classes are represented in the set of pointsthat we have not ruled out. If we include the condition y < q + 1, the set ofpoints remaining is equivalent to the set represented in Theorem 10.19 whichcontains members of all qs+r congruence classes. Since we have ruled out thepoints that satisfy both x ≥ q and y ≥ r, and the points where y ≥ qm + 1,the only points excluded by the condition y < q + 1 will be elements of theBi where i > 0. Thus every residue class represented by an element in the Bi
where i ∈ [1, m− 1] has exactly two representatives in the set of points thatwe have not ruled out. This leaves us with the residues that are not in the Ai
or Bi. If we exclude the Bi where 0 < i < m we get the set of points definedin Theorem 10.19. In the Theorem 10.19 setup each residue is representedonce. Thus residues not represented in the Ai and Bi are represented onceeach in the set of points that we have not excluded.
Also, we should note that all these points are less than αi for i ∈ [1, m]and βi for i ∈ [0, m− 1]. This is because all these points are not in the Ai orBi and are not in the set (x, y)|x ≥ q, y ≥ r, and so these points all havex-values less than q and y-values less than p.We consider all minimal residue systems, that is sets of |A| elements, eachrepresenting a different residue class and each minimal in its residue class.We put the minimal residue systems into m subcollections, which we labelthe 1st, 2nd 3rd... mth subcollection. A minimal residue system is in the ith
subcollection when the βi−1 is an element of that minimal residue system andfor any integer n larger than i-1, βn is not an element of that minimal residuesystem. Note that a minimal residue system cannot contain both αj and βj
for any j, since those two elements are congruent. Consider all the least upperbounds of the minimal residue system in the ith subcollection. We claim thatthere is a minimum least upper bound, and that is lub(αi, βi−1). This valueis achieved in the residue system that contains no element of A1, A2, . . . , Ai−1
but contains every element of Ai, Ai+1, Ai+2, . . . , Am. Every minimal residuesystem in this class contains the points βi−1, and αi since they do not containβi. Thus the minimum possible lub is at least lub(αi, βi−1). This is sufficientto show that the minimum possible lub is lub(αi, βi−1) for this subcollection.
43
If we work in two dimensions, we can do a little better. By a theoremproved last summer, the lub of any minimal system is equal to v + A, wherev is a complete vector. Thus lub(αi, βi−1)-A is a complete vector for alli ∈ 0, 1, ...,m − 1. Thus the Frobenius vectors are the minimal vectors inthis collection.
Theorem 10.21. Given the conditions in Theorem 10.20, and if there aretwo A-vectors, a1 and a2, the Frobenius Vectors are lub((s−1−(i−1)r)b1 +(q−1)b2, (r−1)b1 + iqb2)−A, where i ranges from δχ to δγ inclusive, wherethe δj are the value of i where lub((s−1−(i−1)r)b1+(q−1)b2, (r−1)b1+iqb2)when expressed as x′1a1 + x′2a2 . . . + x′n, has the minimum x′j-value. δχ andδγ are the largest and smallest δjs respectively.
Express b1 as∑n
i=1 κiai and φ2 as∑n
i=1 qiai, where the pi and qi are posi-tive reals. Translate all the points αi and βi by α′
i = αi− (r−1)b1− (q−1)b2
and β′i = βi−(r−1)b1−(q−1)b2. Since we’re translating the points the same
way, the ordering of the lubs remains the same, and in particular the minimallubs remain the same. Express all the lubs in the form
∑ni=1 x′iai. The aj-
value of lub(α′i+1, β
′i), that is, δj is equal to max((s−r)κj−irκj, φj+iqφj). As
i increases from 0 to m, this value is either monotonically increasing, mono-tonically decreasing, or monotonically decreasing and then monotonicallyincreasing. Label the ai such that the deltai are monotonically increasing.Thus, every lub(α′
i+1, β′i) for i < δ1 has all the a1, a2, . . . an coordinate-values
greater than lub(α′δ1+1, β
′δ1
) and every lub (α′i + 1, β′
i) for i > δn has allthe coordinate values greater than lub(α′
δn+1, β′δn
). Given any k, j whereδ1 ≤ k < j ≤ δγ, lub(α′
k+1, β′k) has a smaller x′- value and a greater y′-
value compared to lub(α′j+1, β
′j). Thus any two elements from the set lub
(α′i+1, β
′i) as i ranges from δχ to δγ, are incomparable. Hence any two ele-
ments from the set lub(αi+1, βi) as i ranges from δχ to δγ are incomparable.Thus the set lub(αi+1, βi)−A)as i ranges from δχ to δγ is the set of Frobe-nius vectors.
The values δχ and δγ can be determined easily. We have to first solve theequation (s−r)m1−irm1 = n1+iqn1. δχ equals either bic or die. Solving theequation (s− r)m2 − jrm2 = n2 + jqn2 will similarly get us two candidatesfor δγ, bjc or dje.
Soli Deo Gloria
44
sb1‖≡ b2
This subsection contains some facts and insights about the Frobeniusproblem in the case where m = n = 2 and sb1 is incomparable to b2.Throughout the article, we will assume WLOG that P1(sb1) < P1(b2), andP2(sb1) > P2(b2).
Preliminary Facts: Recall that for 1 ≤ i ≤ b|A|/sc+ 1, we know that theith row of the Selmer lattice is: (i∗s, i∗s+1, i∗s+2, ..., |A|−1). In addition,since sb1 is incomparable to b2, we know that these first b|A|/sc + 1 rowsare in MIN . This is because the zeroes |A| − i, i| where i ranges from 0 tob|A|/sc are all mutually incomparable, and also using Theorem 10.5
Some notation: Let li denote the last element in the ith row of the Selmerdiagram. Let Si denote the set of all elements in row i or above.
Theorem 10.22. For any i such that the ith row of the Minimal SelmerDiagram is non-empty, ∀v ∈ Si, P1(li) ≥ P1(v).
Proof. We induct on i. The base case when i = 1 is clearly true. Assumethe claim is true for i = k, and we will prove it for i = k + 1. It suffices toshow that P1(lk+1) ≥ P1(lk). Here we split the proof into two cases.
Case 1: For 1 ≤ k ≤ b|A|/sc, we have that lk+1 − lk = b1 − sb2, and soP1(lk+1 − lk) = P1(b1 − sb2) ≥ 0 by our assumption that P1(sb1) > P1(b2).
Case 2: For k > b|A|/sc: Recall that row b|A|/sc + 1 of the Selmerdiagram has |A| − b|A|/sc ∗ s < s rows, so row b|A|/sc + 1 contains fewerthan s elements. By (reference Darren’s paper), we know that row k has≤ s elements. This means that lk+1 − lk = b1 − mb2, where m < s. So,P1(lk+1 − lk) = P1(b1 − mb2) ≥ P1(b1 − mb2) ≥ 0. This completes theinduction.
Theorem 10.23. For any i such that the ith row of the Minimal SelmerDiagram is non-empty, there is a unique smallest vector = lub(ω1, ..., ω|A|),where ω1 = li and the set of ω’s are all in the first i rows of the MinimalSelmer Diagram.
Proof. By Property 1, we have that P1(lub(ω1, ..., ω|A|) = P1(li). In addition,there must be a unique minimal value for lub(ω1, ..., ω|A|). This implies thedesired result.
45
Theorem 10.24. For any Frobenius vector g, let lub(ω1, ..., ω|A|)−A1 be aminimal representation of g, where ωi is defined as usual, and let Low(g) =k. Then lk = ωi for some i.
Assume the contrary. Let S = ωi. Let mk be the last element of rowk. WLOG, let ω1 be the element in S that is congruent to lk. We know thatω1 is in a smaller row than lk, and so it must be in a bigger column since ω1
and lk are incomparable. Thus, every element in row k is congruent to someelement in the same row as ω1 and in a smaller column than ω1.
But this allows us to represent g using only the first k−1 rows: For everyωi in the kth row, ∃ an element ω
′i in the same row as ω1, where ω
′i < ω1.
We form S′
in the following way: for each element ωi ∈ S, if ωi is not inthe kth row, then ωi ∈ Si; if ωi is in the kth row, then ω
′i ∈ Si. Notice that
lubS′ ≤ lubS. This cannot be a strict inequality, since g is assumed to be
a Frobenius vector. So, we have that lubS′= lubS, but this contradicts the
fact that S is a minimal representation of v.Consequences of the above three properties: For a given k, let mk,j denote
the minimal P2(ωj), where ωj ≡ j, and lies within the first k rows of theMinimal Selmer Diagram.
Each Frobenius vector can be represented in the following manner: P1(li)a1+minmi,ja2, where j through all congruence classes mod |A| except for thecongruence class of li.
Comment: Note that this also proves that the number of Frobenius vec-tors ≤ the number of rows in the Minimal Selmer diagram, which is ≤ |A|.Theorem 10.25. Let |A| = qs + r. Suppose that sb1‖≡ b2 and that (q +1)b2 ≥ (s−r)b1. Then the Frobenius set is exactly: (|A|−1, 0)b−A1, (|A|−s − 1, 1)b − A1, ..., (|A| − s(b|A|/sc − 1) − 1, b|A|/sc − 1)b − A1, lub((|A| −s(b|A|/sc)− 1, b|A|/sc)b, (|A| − s(b|A|/sc − 1)− r − 1, b|A|/sc − 1)b −A1).
By the wavy cut theorem, we know that in this case the Minimal SelmerLattice is exactly:
0 1 ... |A| − 1s s + 1 ... |A| − 12s 2s + 1 ... |A| − 1...|A| − r − s ... |A| − 1|A| − r ... |A| − 1
46
We know that the elements congruent to |A| − 1 in all but the last roware good, which implies (|A| − 1, 0)b, (|A| − s − 1, 1)b, ..., (|A| − s(b|A|/sc −1) − 1, b|A|/sc − 1)b are all Frobenius vectors. The only remaining elementcongruent to |A|−1 is (|A|−s(b|A|/sc)−1, b|A|/sc)b. Now, we assume WLOGthat P1(sb1 ¡ P1(b2). This means that P1((|A| − s(b|A|/sc)− 1, b|A|/sc)b) ¿P1(v) for any v ∈MIN ≡ |A|−1. Furthermore, it also means that P2((|A|−s(b|A|/sc − 1)− r− 1, b|A|/sc − 1)b) < P2(u) for any u ∈MIN ≡ |A| − r−1. Finally, by 10.32 we know that (|A| − s(b|A|/sc) − 1, b|A|/sc)b, (|A| −s(b|A|/sc − 1)− r − 1, b|A|/sc − 1)b is lub-complete. From this the desiredresult follows.
10.2.2 Convexity of MIN(z)
Note: The following definition and Lemma about the convexity of MIN(z)was originally used in proving Vadim’s conjecture, but they are no longerneeded for that. However, I included it here because it is somewhat interest-ing by itself.
Definition 10.26. For any set of incomparable points in the Selmer Lattice(not necessarily the Minimal Selmer Lattice), we can write the points as(ci, di)b|i ∈ [1, k], where ci > cj iff i < j (from which it follows that di < dj
iff i < j) (note: we let Pi denote (ci, di)b). We define such a set to be convex
when d1−d2
c1−c2> d2−d3
c2−c3> ... > dk−1−dk
ck−1−ck.
Lemma 10.27. For each z ∈ [0, |A| − 1], MIN(z) is a convex set.
Proof. Assume the contrary. Then there exists i such that di−di+1
ci−ci+1< di+1−di+2
ci+1−ci+2.
We assume WLOG that i = 1.We will show that there exists a vector (p, q)b ≡ z mod |A|, with p, q ∈ N0,
c3 < p < c2, and d1 < q < d2. But this will contradict our initial assumptions,because that (p, q)b must be in the Selmer lattice because (p, q)b <B (c2, d2)b,which we assume is in the Selmer lattice, and (p, q)b >B (0, 0)b, so therefore(p, q)b must lie in the Selmer lattice. Furthermore, since (p, q)b <B (c2, d2)b,we have that (c2, d2)b cannot be in MIN , which contradicts our originalassumption.
To simplify the calculations we are about to do, we define a new coor-dinate system (x, y)b′ , where the point (x, y)b = (x − c3, y − d1)b′ . ThenP1 = (c1 − c3, 0)b′ , P2 = (c2 − c3, d2 − d1)b′ , and P3 = (0, d3 − d1)b′ . Finally,let a = c2 − c3, b = d2 − d1, x = (c1 − c2), and y = d3 − d2. So now we
47
have that P1 = (a + x, 0)b′ , P2 = (a, b)b′ , and P3 = (0, b + y)b′ . Finally, letu = P1 − P2 = (a,−y) and v = P2 − P3 = (x,−b). Notice that u ≡ v ≡ 0mod |A|.
The condition that d1−d2
c1−c2< d2−d3
c2−c3translates to −b
x< −y
a, or equivalently
xy < ab. Showing the existence of the vector (p, q)b as defined above trans-lates to finding a vector (r, s)b′ ≡ z mod |A|, with r, s ∈ N0, r < a and s < b.Consider vectors of the form cu + dv + (a, b)b′ , where c, d ∈ Z. All of thesevectors are congruent to z. Since every vector congruent to z mod |A| canbe expressed in this form, finding the vector (r, s)b′ described above is equiv-alent to finding c, d such that cu + dv + (a, b)b′ satisfies the same conditionsas (r, s)b′ . So, it suffices to find integers c, d such that 0 ≤ ca + dx + a < aand 0 ≤ −cy − db + b ≤ b, or equivalently:
−a ≤ ca + dx < 0 (4)
and
0 ≤ cy + db ≤ b (5)
Solving equation (1) gives
c = −bdx
ac − 1 (6)
and equation (2) is satisfied when
d = −bcybc (7)
(note this is not always the unique solution, but this turns out not to matter).So in order to find (r, s)b′ , it suffices to find c, d that satisfy equations (3) and
(4) simultaneously. Substituting d in equation (3), we get c = −b−bcybcx
ac−1,
or equivalently
c = bb cy
bcx
ac − 1 (8)
.When c = 0, the LHS of equation (5) = 0, and the RHS = −1, so in
particular, the LHS > the RHS. When c = −ab, the RHS = xy − 1, so inparticular the LHS ≤ RHS. Notice that as c decreases by 1, the RHS is non-increasing and the LHS obviously decreases by 1. This means that for some
48
0 > c ≥ −ab, we must have that the LHS and the RHS are exactly equal forthat value of c.
10.2.3 Vadim’s Theorem
In this article, we focus on the special case on the vector Frobenius problemwhen n ≥ 2 and m = 2. We also assume that no b vector is parallelto any a vector. The main result is that for every element v ∈ MIN ,∃ a Frobenius vector g and |A| − 1 elements ω1, ω2, ...ω|A|−1 such thatlub(v, ω1, ω2, ...ω|A|−1) − A1 = g. However, in my opinion, the lemmas anddefinitions used in proving this are just as interesting as the result itself.
Definition 10.28. We call an element of the Selmer lattice v a corner pointif v + b1 and v + b2 are both not in the Selmer lattice. We call an elementu ∈MIN a minimal corner point if u+ b1 and u+ b2 are both not in MIN .
Note: recall that we proved in the midterm report that for every vector vthat is both a corner point and a minimal corner point, v−A1 is a Frobeniusvector.
Definition 10.29. We call a set of points ai|i ∈ [1, k] and ai ∈MIN lubcomplete if for every 0 ≤ j ≤ |A|−1 (j ∈ N0), there exists an element vj ≡ jmod |A| − 1 such that vj ∈MIN and vj ≤ ai for some i, not necessarily thesame i for each j.
Definition 10.30. For c1, c2, ...cn ∈ N0, let (c1, c2, ...cn)a denote c1a1+c2a2+... + cnan.
Definition 10.31. For d1, d2 ∈ N0, let (d1, d2)b denote d1b1 + d2b2.
Lemma 10.32. Let r be the largest natural number such that there exists anelement of MIN with b1-value = r, and no element of MIN has a greaterb1-value. Let (r, s) be the element in MIN with b1-value = r, and withthe largest b2-value. Let the significant zero with the greatest b1-value < rbe (x1, y1)b, and let the significant zero with the greatest b1-value < x1 be(x2, y2). Let (p, q) be element of MIN ≡ (r, s) with the largest b1-value notexceeding r. Then the set (p, q − 1), (r, s) is lub-complete.
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(Note: we make the assumption here that both (x1, y1)b and (x2, y2)b, butthis is a harmless assumption because the case where one or both of them donot exist has been completely solved in Darren’s article ”Generalization ofthe first two steps of the Selmer algorithm,” and the main theorem we wishto prove is clearly true in those two cases).
Proof. We first show that p = x1 − 1. By the theorem that every cornerpoint is complete, it follows that there exists an element congruent to (r, s)b
with b1-value ≤ x1−1 and y1 ≤ b2-value ≤ y2−1. Now, we cannot have thatp < x1− 1, because then (p + 1, q)b is in MIN , which implies that (r + 1, s)b
is in MIN (since (r + 1, s)b − (p + 1, q)b = (r, s)b − (p, q)b.Since (x1, y1)b ≡ 0, we know (−x1, 0)b ≡ (0, y1)b. So, it follows that
(r − k ∗ x1, s) ≡ (x1 − 1, q + k ∗ y1), where k ∈ N. This means that everyelement satisfying 0 ≤ b1-value ≤ x1 − 1 and q ≤ b2-value ≤ y2 − 1 iscongruent to some element satisfying 0 ≤ b1-value ≤ r and 0 ≤ b2-value ≤ s,that is, some element in the contained set of (r, s). Recall that the element(x1 − 1, y2 − 1) is a corner point, and therefore it is good. From this thedesired result follows.
Lemma 10.33. Let b1 = (x1, x2, ..., xn)a, and let b2 = (y1, y2, ..., yn)a, andassume WLOG that x1
y1= maxxi
yi and xn
yn= minxi
yi. Then for any two
distinct congruent elements u = (c1, c2)b,v = (d1, d2)b ∈ MIN , d1 > c1 ⇒P1(v) < P1(u) and Pn(v) > Pn(u).
Proof. This follows easily from Darren’s vector division theorem.
Theorem 10.34. For every element v ∈MIN , ∃ a Frobenius vector g and|A| − 1 elements ω1, ω2, ...ω|A|−1 such that lub(v, ω1, ω2, ...ω|A|−1)− A1 =g.
Proof. If an element v ∈ MIN has the desired property (that is, it is usedin some lub), we call v usable. It is easy to see that if v is usable, then anyvector ≤B to v is also usable.
By Darren’s theorem of wavy cuts, we know that all but at most two min-imal corner points are also corner points. If we can show that the remaining
50
minimal corner points (i.e. those that are not also corner points) are usable,then the proof is complete, since every element of MIN is ≤B some minimalcorner point.
Notice that for an element to be a minimal corner point but not a cornerpoint, it must lie on one of the two wavy cuts. We will only deal with thecase where it lies on the vertical wavy cut, as the other case is analogous.Call this element v, and let it be congruent to k mod |A|.
Since v lies in the rightmost column, as a result of Lemma 10.33, we mayWLOG assume P1(v) < P1(u) for any u ∈ MIN(k), u 6= v. Let w be theelement in MIN(k) in the second largest column. By Lemma 10.32 , weknow ∃y such that v,y is lub complete and P1(lub(v, z)) < P1(w) (Noticethat we need the assumption bi is not parallel to aj here to get the strictinequality. Since every P1 of any lub complete set must be ≥ P(z), we haveshown that v is usable.
10.2.4 A bound for the cardinality of the MIN set in higher di-mensions
For n ≥ 2, the following theorem establishes a bound on the cardinality ofMIN(a) (the elements in MIN belonging to the same congruence class) asa function of |A| and m:
Theorem 10.35. For any a ≤ |A|, |MIN(a)| ≤(|A|+ m− 2
m− 1
).
Proof. Recall that all elements of MIN can be represented as∑m
i=1 cibi
where ci ∈ N0 and∑m
i=1 ci ≤ |A| − 1. Given such an element∑m
i=1 aibi
of MIN , define the k-column, where k ∈ [0, m], to be the set c1b1 + ... +ck−1bk−1 + akbk + ck+1bk+1 + ... + cmbm|ak ∈ [0, |A|]
Notice that all the elements of any k-column are comparable to each other.This means that in any k-column, there cannot be two congruent elementsbelonging to MIN . It is also easy to see that every element in MIN belongsto some k-column.
We will count the total number of possible m-columns. Notice that eachm column can be uniquely represented by the numbers ci, where ci ∈ N0.Furthermore, we must have that
∑m−1i=1 ci ≤ |A| − 1. It is well known that
51
the number of ways to choose ci under these conditions is
(|A|+ m− 2
m− 1
).
The desired result follows.
Notice that when m = 1, this bound equals 1. This is reassuring, sincewe know that when m = 1, there is a unique Frobenius vector. Also, whenm = 2, this bound equals |A|, which has been previously proven.
Theorem 10.36. When n = m, this bound is always achievable.
Proof. When all bi are congruent mod |A| (say WLOG they are all congru-ent to 1), then all elements of the set S =
∑mi=1 cibi|
∑mi=1 ci = |A| − 1
are congruent to |A| − 1, and furthermore there are exactly
(|A|+ m− 2
m− 1
)elements in this set. If we can choose bi so that all the elements in S areincomparable, then every element in S must be in MIN , and we would have
|MIN(|A| − 1) =
(|A|+ m− 2
m− 1
).
Define the map Pi(v) ∀i ∈ [1, m] which projects v onto the ai axis. Pickbi such that Pi(bi) > |A| ∗ maxi Pi(bk)|k 6= i. We know we can alwaysdo this because given any vector v, v + ai ≡ v. Now, for any two elementsu and w ≡ |A| − 1, we can write their difference as u − w =
∑mi=1 dibi,
where∑m
i=1 di = 0 and∑m
i=1 |di| ≤ |A|. Assuming u and w are not equal,we must have at least one i such that |di| > 0, and therefore we must havesome i, j such that di > 0 and dj < 0. Then it follows that Pi(u − w) > 0because Pi(dibi) ≥ Pi(bi) ≥ |A| ∗maxiPi(bk)|k 6= i ≥ Pi(
∑mk=1,k 6=i |dkbk|).
Similarly, we have that Pj(u−w) < 0. So, every two elements in S are alsoin MIN , and from this the desired result follows.
Corollary 10.37. Given any n x n matrix A, it is possible to find m b-
vectors which yield exactly
(|A|+ m− 2
m− 1
)different Frobenius vectors.
Proof. We will show that using the same construction for bi as in the prooffor 10.36, each vector in the set S (defined in the proof for theorem 2) isgood. For any v ∈ S, v =
∑mi=1 cibi where
∑mi=1 ci = |A| − 1, define the
set Sv = ∑m
i=1 dibi|di ≤ ci. It is clear that all elements of Sv are lessthan v in cone ordering. Also, since bi ≡ 1∀i, we have that
∑mi=1 dibi ≡
52
∑mi=1 di, and thus this set contains at least one element in each equivalence
class mod |A|. This means that each element of S is good and thereforeeach such element subtracted by A1 is a Frobenius vector, yielding exactly(|A|+ m− 2
m− 1
)different Frobenius vectors.
11 Induced Cone Ordering and lub-complete
vectors
11.1 Definitions
Definition 11.1. Call a vector lub-complete if it is greater than a completeset of coset representatives. Call a minimal lub-complete vector with respectto cone ordering minimal lub-complete. The set of all minimal lub-completeis G +A1 and for each minimal lub-complete vector there is a correspondingg-vector which is less than it by exactly A1.
Definition 11.2. Call vs a supporting vector of lub-complete v with respectto ≤-ordering when vs ≤ v, vs 6≺ v and 6 ∃v′
s ≡ vs with v′s ≤ v.
Definition 11.3. Call two vectors of the same congruence class neighboringif there is no vector of their congruence class between their lub and glb.
Definition 11.4. Denote ≤-relation with respect to B-basis as ≤B
Definition 11.5. Define B-plane to be the plane spanned by B-vectors.
11.2 Induced cone ordering for m = 2
In the same way as A-vectors define cone ordering in Rn, there exist twovectors defining the induced cone ordering on the B-plane. If B-vectors arecolinear, cone ordering defines a total ordering on BN0 . Here we are assumingthat B-vectors are not colinear.
Definition 11.6. On the B-plane define induced cone ordering to be theordering equivalent to the cone ordering on the B-plane. Below we find someproperties of this ordering. This ordering enables us to use arguments fromthe 2-dimensional case in n-dimensions.
53
Below we find the vectors defining the induced cone ordering:
First defineb2,i max
b1,i max= max(
b2,i
b1,i) and
b2,i min
b1,i min= min(
b2,i
b1,i) where i can only
be such that b1,i 6= 0. Analogously defineb1,i max
b2,i maxand
b1,i min
b2,i min.
1. If for all i, b1,i 6= 0 or both b1,i = b2,i = 0, then the induced cone ordering
is defined by the vectors a1 =b2,i max
b1,i maxb1−b2 and a2 = − b2,i min
b1,i minb1 +b2.
2. If for all i, b2,i 6= 0 or both b1,i = b2,i = 0, then the induced cone ordering
is defined by the vectors a1 =b1,i max
b2,i maxb2−b1 and a2 = − b1,i min
b2,i minb2 +b1.
3. If there exist i and j with b1,i = 0, b2,i 6= 0 and b2,j = 0 and b1,j 6= 0,then the induced cone ordering is defined by the vectors a1 = b1 anda2 = b2.
Definition 11.7. Denote ≤-relation defined by the induced cone ordering as≤∼
and ≺-relation as ≺∼. Henceforth A will refer to a1 and a2.
Lemma 11.8. If for all i, b1,i 6= 0 or both b1,i = b2,i = 0, thenb2,i max
b1,i max6=
b2,i min
b1,i min. It then follows
b2,i max
b1,i max6= 0.
Proof. If we suppose the opposite, we haveb2,i0
b1,i0=
b2,i
b1,i(assuming that there
is i0 with b1,i0 6= 0) which means that b2,i =b2,i0
b1,i0b1,i for all i with b1,i 6= 0.
The equality still holds when b1,i = 0 as then we also have b2,i = 0 from theassumptions of the lemma. Thus, we reach contradiction as B vectors areassumed to be not colinear.
Lemma 11.9. In all cases b1, b2 ≥∼
0
Proof. When a1 = b1 and a2 = b2 the claim is obvious. When a1 =b2,i max
b1,i maxb1 − b2 and a2 = − b2,i min
b1,i minb1 + b2 we see that
b1 =a1 + a2
b2,i max
b1,i max− b2,i min
b1,i min
≥∼
0
54
and (as by Lemma 11.8b2,i max
b1,i max6= 0 and
b2,i max
b1,i max>
b2,i min
b1,i min)
b2 =a2 +
b1,i max
b2,i max
b2,i min
b1,i mina1
1− b1,i max
b2,i max
b2,i min
b1,i min
≥∼
0
as claimed.
Theorem 11.10. The induced cone ordering and the cone ordering are equiv-alent on the B-plane.
Proof. It is sufficient to show that ≥ 0 is equivalent to ≥∼
0.
For every c1b1 + c2b2 for it to be ≥ 0 it is necessary and sufficient tosatisfy c1b1,i +c2b2,i ≥ 0 for all i. Next, in separate cases we obtain more spe-cific criteria: If b1,i = 0 and b2,i = 0 the inequality holds trivially. If b1,i 6= 0
it is equivalent to c1 ≥ −c2b2,i
b1,iwhich is in turn equivalent to c1 ≥ −c2
b2,i min
b1,i min
when c2 ≥ 0 and c1 ≥ −c2b2,i max
b1,i maxwhen c2 ≤ 0. If b1,i = 0 and b2,i 6= 0 it is
equivalent to c2 ≥ 0. Finally, if b2,i = 0 and b1,i 6= 0 it is equivalent to c1 ≥ 0.
We need to show that d1a1 + d2a2 ≥ 0 iff d1, d2 ≥ 0 where a1 and a1 areas defined above.
First case: there doesn’t exist i with b1,i = 0 and b2,i 6= 0.
We rewrite d1a1 + d2a2 = (d1b2,i max
b1,i max− d2
b2,i min
b1,i min)b1 + (d2− d1)b2 = c1b1 +
c2b2. Now let’s check when our criterion is satisfied: If d2 − d1 = c2 ≥ 0,c1 ≥ −c2
b2,i min
b1,i minis equivalent to d1
b2,i max
b1,i max−d2
b2,i min
b1,i min≥ −(d2−d1)
b2,i min
b1,i minequiva-
lent to d1(b2,i max
b1,i max− b2,i min
b1,i min) ≥ 0 by Lemma 11.8 equivalent to d1 ≥ 0 which with
d2−d1 ≥ 0 implies d2 ≥ 0. If d2−d1 = c2 ≤ 0, c1 ≥ −c2b2,i max
b1,i maxis equivalent to
d1b2,i max
b1,i max− d2
b2,i min
b1,i min≥ −(d2 − d1)
b2,i max
b1,i maxequivalent to d1(
b2,i max
b1,i max− b2,i min
b1,i min) ≥ 0
which is analogous to the d2 − d1 ≥ 0 case.
Second case: there doesn’t exist i with b2,i = 0 and b1,i 6= 0. This case isanalogous to the first case.
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Third case: there exist i and j with b1,i = 0, b2,i 6= 0 and b2,j = 0, b1,j 6= 0.
In this case d1a1+d2a2 = d1b1+d2b2 ≥ 0 iff c1 = d1 ≥ 0 and c2 = d2 ≥ 0as claimed.
Lemma 11.11. For any v on the B-plane all of its supporting vectors withrespect to the induced cone ordering are also its supporting vectors with respectto the cone ordering.
Proof. Let vs be a supporting vector of v with respect to the induced cone
ordering: vs ≤∼
v, vs 6≺∼
v and 6 ∃v′s ≡ vs, v
′s ≤
∼v. If vs = A
[vs,1
vs,2
]and
v = A
[v1
v2
], then either vs,1 = v1 or vs,2 = v2. As the cases are analogous
let’s deal with vs,1 = v1. When for all i, b1,i 6= 0 or both b1,i = b2,i = 0,
vs,1 = v1 means v − vs is a multiple of a2 = − b2,i min
b1,i minb1 + b2 which implies
that v − vs has a zero coordinate in the i-th dimension where i is the oneat which − b2,i
b1,ireaches its minimum. Cases when for all i, b2,i 6= 0 or both
b1,i = b2,i = 0 and when there exist i and j with b1,i = 0, b2,i 6= 0 and b2,j = 0and b1,j 6= 0 are analogous.
11.3 Clustering of minimal lub-complete in n-dimensions
Definition 11.12. Define lub-cluster to be a minimal lub-complete vectorwith respect to the induced cone ordering.
The following lemma justifies the choice of the term lub-cluster:
Lemma 11.13. For every lub-cluster v there exists a minimal lub-completew with w ≤ v and for every such w its set of supporting vectors includes theset of supporting vectors of v.
Proof. As the v is minimal lub-complete with respect to the induced coneordering it is lub-complete with respect to the cone ordering. Then thereexists a minimal lub-complete w with w ≤ v.
Suppose there exists a supporting vector vs of v which isn’t a supportingvector of w. If vs 6≤ w, then there exists a MIN vector ws with ws ≡ vs
56
and ws ≤ w ≤ v which contradicts with vs being a supporting vector ofv. If vs ≺ w ≤ v, we have vs ≺ v which is in contradiction with vs beinga supporting vector of v. Thus, we can only have vs ≤ w while vs 6≺ w.Also by previous argument we cannot have ws ≡ vs and ws ≤ w, so vs is asupporting vector of w.
Lemma 11.14. There exists a sequence of lub-clusters li = A
[li,1li,2
](i =
1, ..., ilast) such that
1. li,1 < li+1,1 for 1 ≤ i ≤ ilast − 1
2. There doesn’t exist an lub-cluster l′=
[l′1
l′2
]with l
′1 < l1,1 or l
′1 > lilast,1
3. There exist two neighboring supporting vectors s1(i) =
4. There exist supporting points s1 and silast of l1 and lilast respectively,such that s1,1 = l1,1, silast,2 = lilast
and 6 ∃s′1, s′
ilastcongruent to them
with s′1,1 < s1,1, s
′ilast,1
> silast,1.
Proof. Choose lmin to be the least number such that there exists a completeset of coset representatives wii=|A|i=1 with wi,1 ≤ lmin. From minimalityof lmin we must have 4. Property 4 for the last li in the sequence can beproved analogously. As the number of residue classes is finite, there exists
max(wi,2) so that
[lmin
max(wi,2)
]is lub-complete. Then there exists an
lub-cluster l1 ≤[
lmin
max(wi,2)
]which also satisfies l1,1 = lmin and 2 by the
choice of lmin.Now let’s define the rest of the sequence by induction. If we have li let’s
define li+1. Let wj is the set of all supporting vectors of li such that wj,2 =li,2 for all wj. If for some of wj there doesn’t exist a neighboring w
′
j ≡ wj
such that w′j,1 > wj,1 we reached the last vector in our sequence because for
57
any li+1 with li+1,1 > li,1 we have li+1,2 < li,2 = wj,2 and we cannot have acomplete set of coset representatives less than li+1. Otherwise there existsthe set w′
j of neighboring vectors with w′j,1 > wj,1. Let li+1,1 = max(w′
j)
and j be the index of one of w′
j for which the max is achieved. Then
[li+1,1
li,2
]is lub-complete, there are no supporting vectors wj with wj,2 = li,2 so there
must exist an lub-cluster li+1 =
[li+1,1
li+1,2
]with li+1,2 < li,2. Then wj and
w′
j are supporting vectors of li and li+1 respectively. Also as wj and w′
j areneighboring, 3 is satisfied and 1 is obviously satisfied by construction of theli sequence.
11.4 Vadim’s Theorem
Here we show an application of the tools developed above in an alternateproof of Vadim’s Theorem.
For the case m = 2 we want to show that for every element w ∈ MIN,w ≤ g + A1 where g ∈ G.
Set b1,i and b2,i be i-th coordinates of corresponding B-vectors with re-spect to the A-basis. As all B-vectors are in the cone spanned by A-vectorsall bj,i are nonnegative. If B vectors are colinear, then the MIN set lies on theline spanned by the B vectors. As B vectors are inside the A cone, cone or-dering defines total ordering on the line spanned by the B vectors. It followsthat |MIN| = |A| and we have a unique minimal lub-complete vector whichis greater than the entire MIN. Therefore, we can assume that B vectors arenot colinear.
Define B =
[b1,1 b2,1
b1,2 b2,2
]as follows: b1 = b1,1a1 + b1,2a2 and b2 =
b2,1a1 + b2,2a2. As all B-vectors are in the cone spanned by A-vectors allbi,j are nonnegative. As B vectors are not colinear we have |B| 6= 0 and byswapping B-vectors if needed we will always have |B| > 0.
We will need the following lemma:
58
Lemma 11.15. Let two vectors v1 = A
[x1
y1
]= AB
[x1 b
y1 b
]and v2 =
A
[x2
y2
]= AB
[x2 b
y2 b
]be such that x1 < x2 and y1 > y2. Then x1b < x2b
and y1b > y2b. If we start from non-strict inequalities the same proof yieldsnon-strict inequalities.
Proof. First
[x1,b
y2,b
]= B−1
[x1
y1
]= 1
|B|
[b2,2x1 − b2,1y1
−b1,2x1 + b1,1y1
]and analo-
gously
[x2,b
y2,b
]= 1
|B|
[b2,2x2 − b2,1y2
−b1,2x2 + b1,1y2
].
As b2,2(x2 − x1) ≥ 0 ≥ b2,1(y2 − y1) we have |B|x1b = b2,2x1 − b2,1y1 <b2,2x2− b2,1y2 = |B|x2b (the inequality becomes strict since one of the b2,2 andb2,1 is non-zero and one of the inequalities above is strict) and as b1,2(x2 −x1) ≥ 0 ≥ b1,1(y2− y1) we have |B|y1b = −b1,2x1 + b1,1y1 > −b1,2x2 + b1,1y2 =|B|y2b. As |B| > 0 the claim follows.
Now let w1 = A
[w1,1
w1,2
]∈ MIN. Suppose for contradiction that w1 isn’t
less than any minimal lub-complete. Because w1 is in MIN, it isn’t greaterthan any lub-cluster or else by Lemma 11.13 it will be greater than someminimal lub-complete which must be greater than some MIN vector of thesame equivalence class as w1 which contradicts w1 ∈ MIN. Our proof willbe broken into two cases: when there exists an i such that li,1 ≤ w1,1 ≤ li+1,1
and when w1,1 < l1,1 (case when w1,1 > lilast,1 is analogous). First we provethe latter case:
Proof. Let g = A
[g1
g2
]= AB
[g1 b
g2 b
]= l1 and gs = A
[gs,1
gs,2
]= AB
[gs,1 b
gs,2 b
]be a supporting vector of g with gs,1 = g1 and 6 ∃g′
s ≡ gs such that g′s,1 < gs,1
(such gs exists by 4 of Lemma 11.14). Now add to w1 enough copies of b1
and b2 so that it remains in MIN while neither w1+b1 nor w1+b2 is in MIN(this is possible as all MIN elements can be expressed as linear combinationsof B vectors with coefficients < |A|). After this w1 will increase and therestill will be no minimal lub-complete vectors greater than it. If w1,1 ≥ g1,we reach the contradiction in the second part of our proof. By Lemma 11.13for any minimal lub-complete v ≤ g, gs is also a supporting vector of v and
59
ls ≤ v. We cannot have w1 ≤ ls as this implies w1 ≤ v which contradictsour assumption about w1. Then from w1,1 < gs,1, we have w1,2 > gs,2:
w1,1 < gs,1 and w1,2 > gs,2 (9)
From this by Lemma 11.15 we have
w1,1 b < gs,1 b and w1,2 b > gs,2 b (10)
If w1 ≡ gs we reach contradiction with the choice of gs as w1,1 < gs,1.
Since g is an lub-complete there is at least one vector w2 = A
[w2,1
w2,2
]=
AB
[w2,1 b
w2,2 b
]∈ MIN such that w2 ≡ w1 and w2 ≤ g. Among all vectors w2
such that w2 ≡ w1 (w2 6= w1) and w2,1 ≤ gs,1 choose the one with minimalw2,1 and denote it w2. Note that our choice of w2 doesn’t anymore requirethat w2 ≤ g.
If w2,1 < w1,1, we must have w2,2 > w1,2. From this by Lemma 11.15 wehave w2,1 b < w1,1 b and w2,2 b > w1,2 b. Vector
w2 + (gs −w1) = AB
[w2,1 b + gs,1 b − w1,1 b
w2,2 b + gs,2 b − w1,2 b
]≥B
0
by (10) and w2,2 b > w1,2 b. By 0 < gs,1−w1,1 < w2,1 +gs,1−w1,1 < gs,1 (by (9)and w2,1 < w1,1) we found that w2+(gs−w1) ∈ BN0 and as w2+(gs−w1) ≡gs we reach contradiction with the choice of gs.
(by w2,1 b ≤ gs,1 b and (11)) is in BN0 while w1,1 +(gs,1−w2,1) < gs,1 (by (11))and gs ≡ w1 + (gs −w2) which again contradicts with the choice of gs.
60
Also if w2,1b > gs,1 b we cannot have w2,2 b ≥ gs,2 b as this would implyw2 ≥ gs while we know that w2,1 < gs,1. So we are left with the case
w2,1 b > gs,1 b and w2,2 b < gs,2 b (13)
Vectors v1 = w1+b2 and v2 = w1+b1 cannot be in MIN by the choice of
w1. Therefore there exist v1 s = AB
[v1,1 s b
v1,2 s b
]and v2 s = AB
[v2,1 s b
v2,2 s b
]such
that v1 s ≡ v1, v1 s v1 and v2 s ≡ v2, v2 s v2. We must have v1,2 b s =v2,1 b s = 0 or else we have contradiction with w1 ∈ MIN as v1 s − b2 ∈ BN0
or v2 s − b1 ∈ BN0 . Let x = v1 s − b2 = AB
[x1 b
x2 b
]= AB
[v1,1 b s
−1
]and
y = v2 s − b1 = AB
[y1 b
y2 b
]= AB
[−1
v2,2 b s
]. As v1s < v1 and v2s < v2 we
also have x, y < w1 while also x ≡ y ≡ w1 ≡ w2.If x1 b ≥ w2,1 b we reach contradiction as vector
x + (gs −w2) = AB
[x1 b + gs,1 b − w2,1 b
x2 b + gs,2 b − w2,2 b
]is in BN0 (by (13) and x1 b ≥ w2,1 b), x+(gs−w2) ≡ gs and x1+gs,1−w2,1 < gs,1
as x1 < w1,1 < w2,1 which contradicts with the choice of gs. Hence
x1 b < w2,1 b (14)
Let’s examine possible cases:
Case when y 6≥ x: If y1 ≥ x1, then we must have y2 < x2 or else y ≥ x.Yet y1 ≥ x1 and y2 < x1 by Lemma 11.15 imply −1 = y1 b ≥ x1 b ≥ 0 so wecannot have y1 ≥ x1. Then the vector
y + (w2 − x) = AB
[y1 b + w2,1 b − x1 b
y2 b + w2,2 b − x2 b
]≥B
0
(by (14) and x2 b = −1) in BN0 , y+(w2−x) ≡ w1 ≡ w2 and y1 +w2,1−x1 <w2,1 as y1 < x1. The parallelogram with congruent vertices x, y, w2, y +(w2−x) must contain a vector congruent to gs as any vector on BZ can betaken modulus the generating vectors of the parallelogram to obtain a vectorinside the parallelogram congruent to it. As all vectors of BZ on the segmentxy are weighted averages of x and y their B coordinates are at least 0 as
61
x1,2 b = y2,1 b = −1 (excepting x and y). Then also all BZ vectors inside theparallelogram have non-negative B coordinates and so are the vectors on thesegments between y and y + (w2 − x) and between x and w2. Also as allvertices of the parallelogram have smaller first coordinates than that of gs,all vectors inside the parallelogram have smaller first coordinates than thatof gs. Then the existing inside the parallelogram vector congruent to gs isin BN0 (it is not congruent to x ≡ y as we proved that w1 6≡ gs) and givesus contradiction with the choice of gs.
Case when y ≥ x and y2 b ≥ w1,2 b: Vector
w2 + (y −w1) = AB
[w2,1 b + y1 b − w1,1 b
w2,2 b + y2 b − w1,2 b
]≥B
0
(by (12) and y2 b ≥ w1,2 b) is in BN0 and is less than w2 (as y ≤ w1) whichcontradicts w2 ∈ MIN.
Case when y ≥ x and y2 b < w1,2 b: Vector
w1 + (x− y) = AB
[w1,1 b + x1 b − y1 b
w1,2 b + x2 b − y2 b
]≥B
0
(as y1 b = −1 and y2 b < w1,2 b) is in BN0 and is less than w1 (as x ≤ y) whichcontradicts with w1 ∈ MIN.
Now for the second part when w1,1 is between li,1 and li+1,1:
Proof. Again let
g1 = A
[g1,1
g1,2
]= AB
[g1,1 b
g1,2 b
]= li
g2 = A
[g2,1
g2,2
]= AB
[g2,1 b
g2,2 b
]= li+1
and
gs 1 = A
[gs 1,1
gs 1,2
]= AB
[gs 1,1 b
gs 1,2 b
]gs 2 = A
[gs 2,1
gs 2,2
]= AB
[gs 2,1 b
gs 2,2 b
]
62
be neighboring supporting vectors of g1 and g2 respectively with gs 1,2 = g1,2
and gs 2,1 = g2,1 (that they exist follows from 3 of Lemma 11.14). We areassuming g1,1 ≤ w1,1 ≤ g2,1.
Since g1 is an lub-cluster (by Lemma 11.13 and by the assumption onw1) there exists a MIN vector w2 ≡ w1 such that w2 ≤ g1. Let’s assumethat w2,1 < w1,1 as the other case is analogous. As w2 and w1 must beincomparable we also have w2,2 > w1,2 and Lemma 11.15 yields w2,1 b < w1,1 b
and w2,2 b > w1,2 b.It was shown in the previous case that we cannot have w1 ≤ gs 2 so
w1,2 > gs 2,2. By Lemma 11.15 this gives w1,1 b ≤ gs 2,1 b and w1,2 b ≥ gs 2,2 b.Vector
w2 + (gs 2 −w1) = AB
[w2,1 b + gs 2,1 b − w1,1 b
w2,2 b + gs 2,2 b − w1,2 b
]≥B
0
(as w1,1 b ≤ gs 2,1 b and w2,2 b ≥ w1,2 b) is in BN0 , congruent to gs 2, w2,1 +gs 2,1 − w1,1 < gs 2,1 and w2,2 + gs 2,2 − w1,2 < w2,2 < gs 1,2 (as w2,1 < w1,1 andw1,2 > gs 2,2). So we have contradiction with the choice of gs 2.
12 Bounding |G|The following theorem finds a bound on |G(V )| where V is an n × (n + 2)matrix.
Theorem 12.1. Let j be the smallest positive integer such that j · b2 canbe written as a non-negative integer linear combination of a1, . . . , an and b1.Then |G(V )| ≤ j.
In the case where 0, b1, and b2 are linear, it has been proven that |G| = 1.Thus we can assume that BR is a plane.
Definition 12.2. Given a vector v ∈ BR we define the values v1, v2, (v)1, . . .,and (v)n to be the real numbers such that v1b1+v2b2 = v = (v)1a1+· · · (v)nan.
Before we prove this theorem, we will prove the correctness of severalgeometric arguments which will be used. To avoid confusion between theorderings, in this section we will use ≤A to denote our cone partial ordering.We will now be algebraically defining several geometric concepts. These nextseveral definitions do not depend on the condition v||Bw, but many of theproofs are simplified by only defining terms for this case.
63
Definition 12.3. Suppose u, v, w ∈ BR and v||Bw. We say u is above line←→vw if u2 − v2 < −v2−w2
w1−v1(u1 − v1), and u is below line ←→vw if the inequality is
reversed, and u is on line ←→vw if we have equality.
From high school algebra, we can see that u is above ←→vw if u and 0 lieon the same side of ←→vw whether we are looking at the Selmer lattice, or thestandard plane. Also, u is above←→vw iff u is above←→wv. The naming of above isbased on the orientation of the Selmer lattice; looking at the standard planewould suggest the opposite definition.
Definition 12.4. Suppose u, v, w ∈ BR and v||Bw. We say u is above linesegment vw if u is above ←→vw, u1 ≤ max(v1, w1), and u2 ≤ max(v2, w2). Sim-ilarly, we say u is below line segment vw if u is below ←→vw, u1 ≥ min(v1, w1),and u2 ≥ min(v2, w2). Additionally, u is on line segment vw if u is on ←→vw,max(v1, w1) ≥ u1 ≥ min(v1, w1), and max(v2, w2) ≥ u2 ≥ min(v2, w2),
This definition is best understood geometrically by looking at the Selmerlattice. We see that u is above vw iff it is above and to the left of somereal point on vw. Now we will algebraically prove the correctness of severalgeometric arguments.
Lemma 12.5. Suppose u, v, w ∈ BR, v||Bw, and u is on or above vw. Thenu ≤A lub(v, w).
Proof. WLOG assume v2 > w2, so that v1 < w1. If u1 < v1, then becauseu2 ≤ max(v2, w2) = v2 we have u <B v thus u <A v ≤A lub(v, w). Now wecan assume u1 is between v1 and w1, and can write u
′= u1−v1
w1−v1w + w1−u1
w1−v1v
as a weighted average of w and v with non-negative coefficients. Thus fori = 1, . . . , n, we have (u
′)i ≤ max((v)i, (w)i). We also have u
′= u1−v1
w1−v1w +
w1−u1
w1−v1v =
u1−v1
w1−v1(w1b1 + w2b2) + w1−u1
w1−v1(v1b1 + v2b2) =(
u1−v1
w1−v1w1 + w1−u1
w1−v1v1
)b1 +
(u1−v1
w1−v1w2 + w1−u1
w1−v1v2
)b2 =(
u1w1−u1v1
w1−v1w1
)b1 +
(u1−v1
w1−v1w2 − u1−v1
w1−v1v2 + w1−v1
w1−v1v2
)b2 =
u1b1 +(v2 + u1−v1
w1−v1(w2 − v2)
)≥B u1b1 + u2b2 = u,
where the inequality is because u is below ←→vw. Finally, we have u ≤B u′ ≤A
lub(v, w), thus u ≤A lub(v, w).
64
Lemma 12.6. Suppose u, v, w ∈ BR, v||Bw, and u is below vw. Then v +w − u is above vw. If u is instead on vw, then v + w − u is also on vw.
Proof. Let u′= v + w − u. We know that u2 > v2 − v2−w2
w1−v1(v1 − u1), v2 =
v2− v2−w2
w1−v1(v1− v1), and w2 = v2− v2−w2
w1−v1(v1−w1). Taking the last two minus
the first, we see that u′2 < v2 − v2−w2
w1−v1(v1 − u
′1). Thus u
′is above ←→vw. Also,
for i = 1, 2 we are given that ui ≥ min(vi, wi), thus u′i = vi + wi − ui =
Now suppose instead that u is on vw. We still have u′i ≤ max(vi, wi),
and are given that u′i ≥ min(vi, wi). We can add the same three relations
as before, except in this case they are all three equalities. Thus u′2 = v2 −
v2−w2
w1−v1(v1 − u
′1), and u
′is on vw.
Definition 12.7. We will define a complete ordering of the vectors, calledthe row ordering, in BN0 where c >row c
′if either c2 > c
′2 or both c2 = c
′2 and
c1 > c′1. Define maxrow(N) to be the element of N maximal with respect to
≤row. Given g ∈ G there are possibly many complete sets of residues N fromBN0 such that g = lub(N)−A1. We call N a minimal row representation ofg if maxrow(N) is minimized.
Lemma 12.8. For all g ∈ G there exists a row minimal representation of gcontained in the Selmer lattice.
Proof. By Theorem 1.6 there exists at least one complete set of residuesN in MIN ⊂ BN0 such that g = lub(N) − A1. Let N be a row minimalrepresentation of g. For each w ∈ N associate a w
′in the Selmer diagram
from the same residue class such that w ≥B w′. Such a w
′will always exist
because if w is significant, it is in the Selmer lattice, and if w is insignificant,we have w ≥B w − z for some zero z. Let N
′be the set of these associates.
Now lub(N)−A1 ≥ lub(N′)−A1 is complete. So lub(N)−A1 = lub(N
′)−A1,
and N′is a row minimal representation for g from the Selmer lattice.
Lemma 12.9. If lub(N1) − A1 ∈ G and for some complete set of residuesN2 ⊂ BN0 we have lub(N2) ≤A lub(N1), then lub(N2) = lub(N1).
65
Proof. For each w ∈ N2 associate a w′ ∈ MIN from the same residue class
such that w ≥A w′. Let N
′2 ⊂ MIN be the set of these associates. Now
lub(N2)−A1 ≥ lub(N′2)−A1 is complete by Theorem 4.1. But lub(N1)−A1
is a minimal complete point, so we have lub(N2) = lub(N1).
Lemma 12.10. Let g and g′ ∈ G. Let N and N
′be minimal row representa-
tions from the Selmer lattice of g and g′respectively. Suppose that maxrow(N)
and maxrow(N′) are in the same row. Then g = g
′.
Proof. By Lemma 12.8, such a N and N′exist. Suppose that N and N
′are
not equal but have k elements in common. We will show that N and N′could
have been chosen to have k + 1 elements in common. Let m = maxrow(N)and m
′= maxrow(N
′). Let v ∈ N\N ′
and let v′ ∈ N
′with v
′ ≡ v. We havev||Bv
′so WLOG assume that v1 > v
′1.
Case 1: m 6= m′. Because m2 = m
′2 we may assume WLOG, that m <B
m′. Let w ∈ N such that w ≡ m
′. Because m
′> maxrow(N) we have m
′ 6= w,and thus m
′||Bw. Now m′2 > w2 and w1 > m
′1. Now consider w
′= w+m−m
′.
Both w′2 = w2 and w
′1 = m1 + (w1 − m
′1) are in N0, so w
′ ∈ BN0 . Alsow′ ≤ w ≤ lub(N). Thus lub(w
′, N\m) ≤ lub(w
′, N) = lub(N). By Lemma
12.9, lub(w′, N\m) = lub(N), which contradicts the row minimality of N .
Case 2: m = m′and v <B m. We can replace v
′with v in N
′without
changing lub(N′), by Lemma 12.9. Now the new N
′and N have k + 1
elements in common. The case where m = m′and v
′<B m is equivalent.
Case 3: m = m′and v
′is above or on←→mv. We have v
′2 ≤ m2 and v
′1 ≤ v1.
Thus v′
is above mv. By Lemma 12.5, v′ ≤ lub(m, v). Now lub(N) =
lub(v′, N) ≥ lub(v
′, N\v), and by Lemma 12.9, lub(N) = lub(v
′, N\v). Now
v′ , N\v and N′have k + 1 elements in common.
Case 4: m = m′and v
′is below ←→mv. We can assume that we are not in
case 2, so v′||Bm. Thus v
′1 > m1 and v
′2 < v2 so v
′is below mv. By Lemma
12.6, m + v − v′is above mv and by Lemma 12.5, m + v − v
′ ≤ lub(m, v).Notice that m ≡ m + v − v
′. Now lub(N
′) = lub(m + v − v
′, N
′) ≥ lub(m +
v − v′, N
′\m) and by Lemma 12.9, lub(N′) = lub(m + v − v
′, N
′\m), whichcontradicts the row minimality of N
′.
Thus whenever N and N′are not equal, we can increase the cardinality
of their intersection. Applying this argument repeatedly, we see that we canchoose N = N
′, and thus g = g
′.
66
Now we will prove Theorem 12.1.
Proof. Let g ∈ G. Let N be a row minimal representation of g. Let m =maxrow N . Suppose for contradiction that m2 ≥ j. We know that thereexist non-negative integers c1, c2, c3 such that jb2 = c1a1 + c2a2 + c3b1. Nowm = m1b1 + m2b2 = c1a1 + c2a2 + (c3 + m1)b1 + (m2 − j)b2. By replacing mwith (c3 + m1)b1 + (m2 − j)b2, we see that N is not row maximal, and thusm2 < j. The possible values for m2 are 0, 1, . . . , j − 1. But no two g-vectorscan share an m2 value by Lemma 12.10. Thus |G| ≤ j.
This bound is based on all of the generating vectors. As a corollary, weimmediately see that |A| alone can bound |G|.
Corollary 12.11. Let m = 2. Then |G| ≤ |A|.
Proof. Addition mod A is a group of order |A|, so |A|b2 ≡ 0, and |A|b2 ≥ 0.Thus |A|b2 ∈ AN0 ⊂ MN0 . Defining j as in Theorem 12.1, we see that|G| ≤ j ≤ |A|.
13 Clustering
Definition 13.1. Define h-vectors to be minimal g-complete vectors withrespect to cone ordering on Zn.
Given coordinates of one g-vector we can find all h-vectors which arestrictly inside its cone.
First we need some classification of lines parallel to a1 or a2 on the lattice.If either a1 or a2 have GCD of their coordinates not equal to 1, we factorthis GCD out and obtain a pair of vectors with relatively prime coordinates.When coordinates are relatively prime, it is easier for us to describe theintegral points on lines parallel to a1 and a2. To these vectors we shallfurther refer as a1 r and a2 r. The matrix composed of these new vectorsshall be referred to as reduced A matrix and denoted Ar. To avoid confusionwith orientation, let a2 r have a steeper slope than a1 r.
Equation of a line parallel to a1 r and passing through an integral point(x1, y1) is a1,2 r(x − x1) − a1,1 r(y − y1) = 0. This line intersects 0y axis atthe point (0, y1 − a1,2 rx1
a1,1 r) = (0, a1,1 ry1−a1,2 rx1
a1,1 r). Thus we can classify each line
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parallel to a1 r by the integer N which we will call id number with respect toorigin.
Line equation with id N can be rewritten as a1,2 rx− a1,1 r(y− Na1,1 r
) = 0.
Then integral points on the line can be described as follows: x = ka1,1 r −a−1
1,2 rN and y = Na1,1 r
+a1,2 r(ka1,1 r−a−1
1,2 rN)
a1,1 r= ka1,2 r + N
1−a1,2 ra−11,2 r
a1,1 rwhere
the inverses are taken modulus a1,1 r (and so we mean in later parts ofthe document; inverse exists as GCD(a1,1 r, a1,2 r) = 1) and k ∈ Z. If wemake the translation of basis to the one defined by reduced A matrix the
points will be described as follows: x = k − Na−11,2 r|Ar|+a2,1 r
a1,1 r|Ar| , y = N|Ar| with
|Ar| = a1,1 ra2,2 r − a1,2 ra2,1 r. Notice that l = −a−11,2 r|Ar|+a2,1 r
a1,1 ris an integer.
Thus x = k + N l|Ar| , y = N
|Ar| !.
Let g =
[g1
g2
]= Ar
[g′1
g′2
]denote a g-vector. It is straightforward to
find the greatest integral point with the same a2 r coordinate as g and lessthan g. First we find the id of the line parallel to a1 r passing through g with
respect to origin: a1,1 rg2 − a1,2 rg1. Denoting the point h = Ar
[h′1
h′2
], h
′1
must be maximal such that h′1 = k + (a1,1 rg2 − a1,2 rg1)
l|Ar| ≤ g
′1. From this
we have k = bg′1 − (a1,1 rg2 − a1,2 rg1)l
|Ar|c so
h = Ar
[h′1
h′2
]= Ar
[bg′1 − (a1,1 rg2 − a1,2 rg1)
l|Ar|c+ (a1,1 rg2 − a1,2 rg1)
l|Ar|
g′2
]Analogously we find the greatest integral point less than g and with the samea1 r coordinate as g:
a2,2 r(this time only the inverse is taken modulus a2,2 r).
Now if we move origin to h for our convenience, it is easy to calculatecoordinates of g with respect to h. Next with respect to this new origin weconsider the lines parallel to a1 r. They will be inside the cone of g if and onlyif their id number is positive as id has the same sign as the a2 r coordinateof points on the line. Also the ids are bounded by the id of the line passingthrough f + a2 r which is exactly (f
′2 + 1)|Ar| = M . As we increase the id N
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while 0 ≤ N ≤ (f′2 +1)|Ar| for each line with the id we find the point suspect
to be an h-vector – the least point on the line which is inside the cone: Recallthat the general form of a1 r coordinate of an integral point on the line withN id is k+N l
|Ar| . If the coordinate is greater than 1, which is the coordinateof h + a1 r, the point cannot be an h-vector. Also the coordinate cannot benegative and still! be inside the cone of g. This leaves the a1 r coordinate onlyto be mod(lN,|Ar|)
|Ar| . To decide whether the point is an h-vector, compare its a1 r
coordinate with the a1 r coordinate of the previous h-vector obtained by thisprocess (if this is the first line we need to compare with the a1 r coordinateof h + a1 r = 1). If it is less than that of the previous and still greater thanthat of f + a2 r = M
|Ar| , it is an h-vector; otherwise it isn’t an h-vector. So ifwe define the sequence of a1 r coordinates of suspect h-vectors multiplied by
|Ar|, mod(lN, |Ar|)N<(f′2+1)|Ar|
N=1 (mod function here returns remainders from1 to |Ar|), the strictly decreasing subsequence obtained from it by choosingonly terms which are less than the previously chosen terms and greater thanM = (f
′1 + 1)|Ar|, will correspond to the h-vectors which are strictly inside
the cone of g. As the number of terms in such sequences is bounded by|Ar|− 2 we can have no more than |Ar|− 2 h-vectors strictly inside the cone.However, |Ar| − 2 is not achievable for all Ar matrices and we can give a
better bound if we use l: First b |Ar|mod(−l,|Ar|)c terms will be decreasing and the
rest must be less than mod(b |Ar|mod(−l,|Ar|)cl, |A
r|) which yields:
Theorem 13.2. Given a reduced matrix Ar, the number of h-vectors con-tained strictly inside a cone of a g-vector is bound by b |A
r|dc+mod(|Ar|, d)−1
where d = mod(a−11,2 r|Ar|+a2,1 r
a1,1 r, |Ar|)
14 Unbound
Definition 14.1. For a vector v let vi denote i-th coordinate with respect tothe standard orthonormal basis and let v
′i = (A−1v)i. For example, for gk,
g′
k,i = (A−1gk)i.
Definition 14.2. Define e-class to mean equivalence class with respect to A.If we have a generating vector b1 we can assign a residue number t(x) (0 ≤t(x) ≤ |A| − 1) to each vector x so that x ≡ t(x)b1 mod(A). This numberwill correspond to an e-class and we shall further refer to e-classes by thesenumbers.
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Definition 14.3. For a vector g, we say that it contains vector x wheng + A1 ≥ x which is equivalent to g
′i + 1 ≥ x
′i for every i. We say that it
contains a residue class when it contains a vector from this residue class.
Theorem 14.4. Let b1 be a generator of all residue classes and b2 be suchthat b2 ≡ tb1 mod(A) (1 ≤ t ≤ |A| − 1) while b
Proof. Define s = |A| − t.From the restriction on k we have k(sb
′1,1 + b
′2,1) ≤ (|A| − s)b
′1,1 or kb
′2,1 +
(s− 1)b′1,1 ≤ (|A| − ks− 1)b
′1,1.
Also as b′2,2 > (|A| − 1)b
′1,2, kb
′2,2 + (s − 1)b
′1,2 ≥ b
′2,2 ≥ (|A| − 1)b
′1,2 ≥
(|A| − ks− 1)b′1,2.
Putting this together we have gk = A
[g′
k,1
g′
k,2
]= lub((|A| − k(|A| − t) −
1)b1, kb2 + (|A| − t− 1)b1)−A1 = A
[(|A| − ks− 1)b
′1,1 − 1
kb′2,2 + (s− 1)b
′1,2 − 1
]First we prove that gk are g-complete.
As b1 is a generator (|A|−1)b1−A1 is g-complete. Also since gk +A1 ≥(|A| − 1 − ks)b1 ≥ (|A| − 1 − ks − i)b1, (0 ≤ i ≤ |A| − 1 − ks), gk
contains the following e-classes: 0, 1, ..., |A| − 1 − ks and as gk + A1 ≥kb2 + (s − 1)b1 ≥ ib2 + jb1, (1 ≤ i ≤ k, 0 ≤ j ≤ s − 1), it also contains|A| − ks, |A| − ks + 1, ..., |A| − (k − 1)s− 1, |A| − (k − 1)s, |A| − (k − 1)s +1, ..., |A| − (k− 2)s− 1, ..., |A| − s, |A| − s + 1, ..., |A| − 1 and is g-complete.(here we are using Theorem 4.1)
Second we prove that gk are minimal g-complete.
For g0, g0 + A1 6≥ ib2 + jb1, (i > 0) as b′2,2 > (|A| − 1)b
′1,2, so g0 =
lub(0,b1, ..., (|A| − 1)b1)−A1 is a g-vector as any vector which is less thang0 will not contain |A| elements from MIN .
For gk, elements (k + i)b2 + jb1 (i > 0) are not contained by gk as(k + i)b
′2,2 + jb
′1,2 ≥ kb
′2,2 + b
′2,2 > kb
′2,2 + (|A| − 1)b
′1,2 > kb
′2,2 + (s− 1)b
′1,2.
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Now if we show that (|A| − ks − 1)b1 and kb2 + (s − 1)b1 are the onlyelements of their respective e-classes that are contained by gk, we will provethat gk is a g-vector.
For i ≤ k elements ib2+jb1 congruent to kb2+(s−1)b1 are not containedby gk because ib2+jb1 ≡ kb2+(s−1)b1 implies j ≡ s(i−k)+s−1 mod(|A|)or j = n|A| + s(i − k) + s − 1 where n > 0 as j ≥ 0 (except when i = k,n ≥ 0). When i < k we have ib
′2,1 + jb
′1,1 ≥ jb
′1,1 ≥ (|A| − ks + s − 1)b
′1,1 >
(|A| − ks− 1)b′1,1 = g
′
k,1 + 1 and gk doesn’t contain ib2 + jb1. When i = kwe have ib2 + jb1 = kb2 + (n|A| + s− 1)b1 ≥ kb2 + (s− 1)b1 (for n > 0)which proves that kb2 + (s − 1)b1 is the only element of its e-class that iscontained by gk.
Analogously, for elements ib2 + jb1 congruent to (|A|−ks−1)b1 are notcontained by gk because ib2 + jb1 ≡ (|A| − ks− 1)b1 implies j ≡ s(i− k)−1 mod(|A|) or j = n|A|+s(i−k)−1 where n > 0 as j ≥ 0. This in turn impliesthat ib
′2,1 +jb
′1,1 ≥ jb
′1,1 ≥ (|A|+s(i−k)−1)b
′1,1 ≥ (|A|−ks−1)b
′1,1 = g
′
k,1 +1(we have strict inequality when i > 0) which shows that (|A| − ks− 1)b1 isthe only element of its e-class that is contained by gk.
Corollary 14.5. If A =
[a1 00 a2
]and b1 =
[11
]we can have max(a1, a2)
g-vectors.
Proof. Use Theorem 1 with t = |A| − 1, b2,1 = a1 − 1, b2,2 = la2 − 1 where lis such that b2,2 > (a1a2 − 1)b1,2.
15 Further Possibilities
On the topic of adjacency:Explore the structure of blocks in n > 2 dimensions.It is not true, as we at one time considered, that all blocks are either of
size n when |G| ≥ n or size |G| otherwise. It may be worthwhile to attempta characterization of these blocks.
On the topic of order and the MIN set:
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Let α(bi) ∈ N be the smallest element such that α(bi)bi ∈ [M − bi]N0 . Wedefine the following set:
S ′M =
m∑
i=1
cibi | 0 ≤ ci < α(bi)
Conjecture 15.1. MIN ⊆ S ′M
This result would be an improvement on Lemma 3.1. Furthermore, con-sider x ∈ Nm such that Bx is minimal and Bx ≡ 0 mod A. Let X be theset of all such x. We denote the ith coordinate in x as (x)i. Now considerthe following set:
S′M =
m∑
i=1
cibi | (x)i ≤ ci < α(bi),∀x ∈ X
Consider this rough argument:Let v ∈ S
′M ⊆ S ′
M . Then v = s + γBx where s ∈ S ′M \ S
′M . Thus
v = s + γBx ≡> s + 0γ
implies that v /∈MIN . Therefore MIN ⊆ S ′M \ S
′M .
The idea is that this set S′M contains the elements in S ′
M (or simply SM)that are not in MIN due to this Bx element. We have the following twopossible conjectures, the first being a weaker version of the second. In then = m = 2 case, the second possibility seems to be true.
Conjecture 15.2.
• MIN ⊆ S ′M \ S
′M
• S ′M \ S
′M = MIN
On the topic of the Selmer Lattice:Extend the Selmer Lattice to m > 2.
It may be possible to extend Theorem 12.1:
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Conjecture 15.3. Let m = 3. Let j1 be the smallest integer such that j1b2
is in [A|b1]N0 and let j2 be the smallest integer such that j2b3 is in [A|b1|b2]N0.Then |G| ≤ j1j2.
Other conjectures:The following conjecture concerns the “shape” of the distribution of the
g vectors in a monoid.
Conjecture 15.4. Elements of G are either concave up (with respect to coneordering) or lie along a line.
